Class 11 Commerce Applied Mathematics - Logarithm And Antilogarithm

Write the characteristic of 1235.3 by using their standard forms:

$1235.3=1.2343\times { 10 }^{ 3 }$
Hence characteristic is 3

Simplify:

$\dfrac{8^{-1} \times 5 ^{3}}{2^{-4}}$

$\frac{8^{-1} \times 5 ^{3}}{2^{-4}}$ =

$\frac{1}{8}\times{5}^{3}\times{2}^{4}$ =

${5}^{3}\times{2} = 250$

If $\displaystyle \log_{10} .001 = a$ , then $-a$ =

Let,

$\log_{10}x = a$

$\implies x = 10^a$
Therefore,

$0 .001 = 10^a$

$\Rightarrow a=-3$ .

Compute:
$(4^7)^2\times (4^{-3})^4$

$(4^7)^2\times (4^{-3})^4$

$=4^{14}\times 4^{-12}$

$=4^{14-12}$

$=4^2$

$=16$

Compute:
$(\dfrac{56}{28})^0 \div (\dfrac{2}{5})^3 \times \dfrac{16}{25}$

As given

$\Rightarrow \left(\frac{56}{28} \right)^0\div\left(\frac{2}{5} \right)^3\times\frac{16}{25}$

$\because \left(\frac{56}{28} \right)^0=1$

$\Rightarrow 1\div\frac{8}{125}\times\frac{16}{25}$

$\Rightarrow \frac{1}{\frac{8}{125}}\times\frac{16}{25}$

$\Rightarrow \frac{125}{8}\times\frac{16}{25}$

$\Rightarrow 5\times2=10$

Compute:
$(2^{-9}\div 2^{-11})^4$

$(2^{-9}\div 2^{-11})^4$

$=(2^{-9-(-11)})^4$

$=(2^2)^4$

$=2^8$

$=256$

Find the characteristic of the logarithm of the number $5395.$

We have to find

$\log 5395$
Logarithm of a number has 2 parts - characteristic and mantissa.
5395 can be written in standard form as

$5.395 \times 10^{3}$
So, the characteristic is 3.

$(12)^{-2} \times 3^2$ is equal to $\dfrac{1}{16}$ .
If true then enter $1$ and if false then enter $0$

$\Rightarrow \left(12 \right)^{-2} \times 3^2$

$\Rightarrow \dfrac{3^2}{12^2}\Rightarrow \dfrac{9}{144}$

$\Rightarrow \dfrac{1}{16}$

Compute the absolute value of:
$(-5)^4 \times (-5)^6 \div (-5)^{9}$

Given that:

$\left(-5 \right)^4\times\left(-5 \right)^6\div\left(-5 \right)^9$

$\because a^m\times a^n=a^{m+n}$
so

$\Rightarrow \left(-5 \right)^{10}\div\left(-5 \right)^9$

$\because a^m\div a^n=a^{m-n}$
so

$\Rightarrow \left(-5 \right)^{10-9}$

$\Rightarrow - 5$

$\therefore$ The absolute value is

$5$

Compute:
$(9^0 \times 4^{1})$

we know that,

$\because a^0=1$

so,

$9^0=1$

we also know that

$4^1=4$

$\implies 9^0*4^1=4$

The value of x,if $\log_{10} x = -2$ is $k$ .
then $100x$ =?

Since,

$\log _{ 10 }{ x } = -2$

$\implies \mbox{In exponent form}$

$x = { 10 }^{ -2 }$

$\implies x = 0.01 = k$

Thus,

$100x=100 \times 0.01 = 1$

Show that $\displaystyle \left ( 2\sqrt{2} \right )^{-2/3}= \frac{1}{2}.$ can be written as $\displaystyle \log _{2\sqrt{2}} 2= -\frac{2}{3}$

Given equation

$\displaystyle \left ( 2\sqrt{2} \right )^{-2/3}= \frac{1}{2}.$

Taking log with base

$2\sqrt2$

$\displaystyle \log _{2\sqrt{2}} (\frac{1}{2})= -\frac{2}{3}$

$\displaystyle \log _{3}\frac{1}{243}= -5.$

Converting into exponential form,

$\displaystyle \frac{1}{243}= 3^{-5}$

$\Rightarrow 3^5=243$

Show that $\displaystyle 5^{0}= 1$ can be written as $\displaystyle \log _{25}1= 0$

$5^0=1$
Taking log with base

$25$

$0 \log _{ 25 }{ 5 } =\log _{ 25 }{ 1 }$

$\log_{25}1=0$

Express $\displaystyle \log _{100}0.1= -\frac{1}{2}$ in exponential form:

Given equation

$\displaystyle \log _{100}0.1= -\frac{1}{2}$

Converting into exponential form,

$\displaystyle 0.1= 100^{\frac{-1}{2}}$

$\Rightarrow 0.1 =(10^2)^{\frac{-1}{2}}$

$\Rightarrow 0.1=10^{-1}$

$\displaystyle 4^{3/2}= 8$

$\textbf{Step 1 : Using definition of logarithm}$

$\boldsymbol{x = a^{b} }$

$\Rightarrow \boldsymbol{\log_{a} x = b}$ ;

$x > 0$

$\, ; a > 0 ,$

$\neq 1$

$\textbf{Given : }$

$4^{3/2} = 8$

$\Longrightarrow\displaystyle \log _{4}8= \frac{3}{2}$

If $3^{-2}$ is $\cfrac{1}{m}$ , then the value of $m$ is

$3^{-2}$

$= \displaystyle \frac{1}{3^2}$

$= \frac{1}{3\times 3}$

$= \frac{1}{9}$

If $(-4)^{-2}$ is $\cfrac{1}{m}$ , then the value of $m$ is

$\displaystyle (-4)^{-2}$

$= \frac{1}{(-4)^2}$

$= \frac{1}{(-1)^2 \times (4)^2}$

$= \frac{1}{1 \times 16}$

$= \frac{1}{16}$

$(3^0 + 4^{-1}) \times 2^2$

$\displaystyle (3^0 + 4^{-1}) \times 2^2$

$= \left ( 1 + \frac{1}{4} \right ) \times 4$

$= \frac{4 + 1}{4} \times 4$

$= \displaystyle \frac{5}{4} \times 4$

$= 5$

Evaluate $\displaystyle \left ( \frac{5}{8} \right )^{-7} \times \left ( \frac{8}{5} \right )^{-4}$ is $\frac { 512 }{m }$
Value of $m$ is

Formulae to be used :

$a^m\times a^n = a^{m+n}$

$a^{-n}=\frac{1}{a^n}$

$(\frac { 5 }{ 8 } )^{ -7 }\times(\frac { 5 }{ 8 } )^{ 4 }\quad \\ =\ \ \ (\frac { 5 }{ 8 } )^{ 4-7 }\\ =\ \ (\frac { 5 }{ 8 } )^{ -3 }\\=\ \ (\frac { 8 }{ 5 } )^{ 3 } \\ =\ \ \frac { 512 }{ 125 }$

Evaluate $(5^{-1} \times 2^{-1}) \times 6^{-1}$ is $\frac{1}{m}$
Value of $m$ is

$(5^{-1} \times 2^{-1}) \times 6^{-1}$

$=\frac{1}{5}\times \frac{1}{2}\times \frac{1}{6}$

$=\frac{1}{60}$

Evaluate $\displaystyle \frac{8^{-1} \times 5^3}{2^{-4}}$

$\displaystyle \frac{8^{-1}\times 5^3}{2^{-4}}$

$= \frac{1}{8} \times 5^3 \times 2^4$

$= \frac{1}{8} \times 125 \times 16$

$=125\times 2$

$=250$

Find the value of $m$ for which $5^m \div 5^{-3} = 5^5$

$5^m \div 5^{-3} = 5^5$

$\Rightarrow \displaystyle 5^m \div \frac{1}{5^3} = 5^5$

$\Rightarrow 5^m \times 5^3 = 5^5$

$\Rightarrow 5^{m + 3} = 5^5$

$\Rightarrow m + 3 = 5$

$\Rightarrow m = 5 - 3 = 2$

Evaluate: $-\displaystyle \left \{ \left ( \frac{1}{3} \right )^{-1} - \left ( \frac{1}{4} \right )^{-1} \right \}^{-1}$

$- \left \{ \left ( \cfrac{1}{3} \right )^{-1} - \left ( \cfrac{1}{4} \right )^{-1} \right \}^{-1}$

$= -(3^1 - 4^1)^{-1}=-(-1)^{-1}$

$= \cfrac{-1}{(-1)^1} = 1$

Find $x$ , if $(4^{-1} + 8 ^{-1} ) \times (3^{-1} - 9^{-1}) \div \displaystyle \frac{1}{12} = 5^x$

$\left( \cfrac{1}{4} + \cfrac{1}{8}\right) \times \left( \cfrac{1}{3} - \cfrac{1}{9} \right) \times \cfrac{12}{1} = 5^x$

$\Rightarrow \left( \cfrac{2+1}{8}\right) \times \left( \cfrac{3-1}{9}\right) \times \cfrac{12}{1} = x^x$

$\Rightarrow \cfrac{3}{8} \times \cfrac{2}{9} \times \cfrac{12}{1} = 5^x$

$\Rightarrow 1 = 5^x$

$\Rightarrow 5^0 = 5^x$

$\Rightarrow x = 0$

By what number should $\displaystyle \left ( \frac{-2}{3} \right )^{-3}$ be divided so that the quotient may be $\displaystyle \left ( \frac{4}{27} \right )^{-2}$

Let the number be

$x$

$\displaystyle \frac{ \left ( \frac{-2}{3} \right )^{-3}}{x}= \left ( \frac{4}{27} \right )^{-2}$

$\displaystyle\Rightarrow\frac{ \left ( \frac{-3}{2} \right )^{3}}{x}= \left ( \frac{27}{4} \right )^{2}$

$\displaystyle\Rightarrow x=\dfrac{(\frac{-3}{2} )^{3}}{(\frac{27}{4} )^{2}}$

$\displaystyle\Rightarrow x=\dfrac{(\frac{-27}{8} )}{(\frac{27^2}{16} )}$

$\displaystyle\Rightarrow x=\dfrac{-2}{27}$

Simplify: $\left(\cfrac{3}{2} \right)^{0} \times \left(\cfrac{4}{5}\right)^{-2}$

$\displaystyle \left ( \frac{3}{2} \right) ^0 \times \left ( \frac{4}{5} \right )^{-2}$

$=1 \times \left ( \cfrac{4}{5} \right )^{-2}$

$= \left ( \cfrac{5}{4} \right )^{2}$

$= \cfrac{25}{16}$

Simplify : $\displaystyle \frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} (t \neq 0)$

$\cfrac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}$

$=\cfrac{5^2 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}$

$=\cfrac{5^{2+3} \times t^{-4+8}}{ 5\times 2 }$

$=\cfrac{5^{5} \times t^{4}}{ 5\times 2 }$

$=\cfrac{5^{4} \times t^{4}}{ 2 }$

$=\cfrac{625\times t^4}{2}$

Simplify: $3^0 +2^{-2}$

${ 3 }^{ 0 }+{ 2 }^{ -2 }= 1 + \cfrac { 1 }{ { 2 }^{ 2 } }$

$=1 +\cfrac { 1 }{ 4 }$

$= \cfrac { 5 }{ 4 }$

Simplify $\displaystyle \left [ \left ( \frac{2}{3} \right )^{-1} \times \left ( \frac{3}{4} \right )^{-1} \right ]^{-1}$

$\displaystyle \left [ \left ( \frac{2}{3} \right )^{-1} \times \left ( \frac{3}{4} \right )^{-1} \right ]^{-1}$

$=\displaystyle \left [ \left ( \frac{2}{3} \right )^{-1 \times (-1)} \times \left ( \frac{3}{4} \right )^{-1\times (-1)} \right ]$

$=\displaystyle \left [ \left ( \frac{2}{3} \right ) \times \left ( \frac{3}{4} \right ) \right ]$

$=\frac{1}{2}$

Find the value $\displaystyle 27^{\cfrac{1}{3} } \times 16^{\cfrac{-1}{4} }$

$\displaystyle 27^{\displaystyle \frac{1}{3} } \times 16^{\displaystyle \frac{-1}{4} }$

$\Rightarrow \left(3^{3}\right)^{\frac{1}{3}} \times \left(2^{4}\right)^{\frac{-1}{4}}$

$\Rightarrow \left(3^{3\cdot \frac{1}{3}}\right) \times \left(2^{4 \cdot \frac{-1}{4} }\right)$

$\Rightarrow 3 \times 2^{-1}$

$\Rightarrow \frac{3}{2}$

By what number should $\displaystyle \left ( \frac{3}{4} \right )^{-3}$ be divided so that the quotient becomes $128$ ?

Let, the number be

$x$

ATQ,

$\displaystyle \left ( \dfrac{3}{4} \right )^{-3} \div x=128$

$\Rightarrow x=\dfrac{\left( \dfrac{3}{4} \right)^{-3}}{128}$

$\Rightarrow x=\dfrac{\left( \dfrac{4}{3} \right)^{3}}{128}$

$\Rightarrow x=\dfrac{4^3}{3^3\times 128}$

$\Rightarrow x=\dfrac{64}{27\times 128}$

$\Rightarrow x=\dfrac{1}{27\times 2}$

$\Rightarrow x=\dfrac{1}{54}$

Hence, the required number is

$\dfrac{1}{54}$ .

Find the value $(512)^{\frac{-2}{9}}$

Given that

Let

$a=(512)^{ \frac{-2}{9}}$

$= \left(2^{9}\right)^{\frac{-2}{9}}$

$= \left(2^{\ 9\times \frac{-2}{9}}\right)$

$= 2^{-2}$

$= \dfrac{1}{2^{2}}$

$= \dfrac{1}{4}$

$\Rightarrow (512)^{ \frac{-2}{9}}= \dfrac{1}{4}$

For what negative values of $x$ , will $x^{18}$ be equal to $x^{20}$ ?

Given,

$x^{20}=x^{18}$

$\Rightarrow x^{20}-x^{18}=0$

$\Rightarrow x^{18}(x^2-1)=0$

$\Rightarrow x^{18}(x+1)(x-1)=0$

$x=0,x=1,x=-1$

Considering only negative values, at

$x=-1$ ,

$x^{18}$ will be equal to

$x^{20}$ .

For what positive values of $x$ , will $x^{18}$ be greater than $x^{20}$ ?

$x^{18}$ will be greater than

$x^{20}$ for

$x>0$ ,

$x=\cfrac{p}{q}$

where,

$p,q \neq 0$ and

$p<q$

For what negative values of $x$ , will $x^{20}$ be greater than $x^{18}$ ?

$x^{20}$ will be greater than

$x^{18}$ ,

$x=-\cfrac{p}{q}$ , where

$p,q \neq 0$ and

$p>q$ .

If $\displaystyle \log_{3}a=4$ , then find the value of $a$ .

Given,

$\log_3a=4$

Converting it in exponential form

Therefore,

$\displaystyle a=3^{4}$

$\Rightarrow a=81$

Find the value of x if $x^3 = \displaystyle \left ( \frac{6}{5} \right )^{-3} \times \left ( \frac{6}{5} \right )^6.$

${ x }^{ 3 }\quad =\quad ({ \dfrac { 6 }{ 5 }) }^{ -3 }{ \times (\dfrac { 6 }{ 5 } ) }^{ 6 }\quad =\quad { \dfrac { 6 }{ 5 } }^{ -3+6\quad }\\ { x }^{ 3 }\quad =\quad ({ \dfrac { 6 }{ 5 } ) }^{ 3 }\\ So,\quad x\quad =\quad \dfrac { 6 }{ 5 }$

${ 3 }^{ 3 }\times { 3 }^{ 6 }\times { 3 }^{ 7 }=$ ?

${ 3 }^{ 3 }\times 3^{ 6 }\times { 3 }^{ 7 }={ 3 }^{ \left( 3+6+7 \right) }={ 3 }^{ 16 }$ [Using

$a^m \times a^n = a^{m+n}$ ]

Suppose $m$ and $n$ are distinct integers. Can $\cfrac { { 3 }^{ m }\times { 2 }^{ n } }{ { 2 }^{ m }\times { 3 }^{ n } }$ be an integer? Give reasons.

Given that:

$\dfrac { { 3 }^{ m }\times { 2 }^{ n } }{ { 2 }^{ m }\times { 3 }^{ n } }$

( m and n are distinct integers)

$=\dfrac { { 3 }^{ \left( m-n \right) } }{ { 2 }^{ \left( m-n \right) } }$ ........Using [

$a^{-m} = \dfrac{1}{a^m}$ ]

$={ \left( \dfrac { 3 }{ 2 } \right) }^{ m-n }$

this cannot be integer as

$\dfrac { 3 }{ 2 } =1.5$ so

${ \left( 1.5 \right) }^{ \left( m-n \right) }$ whether

$m>n$ or

$n>m$ in all cases result will be decimal. So

$\dfrac { { 3 }^{ m }\times { 2 }^{ n } }{ { 2 }^{ m }\times { 3 }^{ n } }$ is not an integer.

${ 2 }^{ 5 }\times { 5 }^{ 2 }\times { 2 }^{ 3 }\times 5=$ ?

${ 2 }^{ 5 }\times { 5 }^{ 2 }\times { 2 }^{ 3 }\times 5={ 2 }^{ \left( 5+3 \right) }\times { 5 }^{ \left( 2+1 \right) }={ 2 }^{ 8 }\times { 5 }^{ 3 }$ [Using

$a^m \times a^n = a^{m+n}$ ]

${ 2 }^{ 5 }\times { 2 }^{ 6 }=2^?$

${ 2 }^{ 5 }\times { 2 }^{ 6 }={ 2 }^{ \left( 5+6 \right) }={ 2 }^{ 11 }$ Using [

$a^m \times a^n = a^{m+n}$ ]

Simplify:
${ 3 }^{ 1 }\times { 3 }^{ 2 }\times { 3 }^{ 3 }\times { 3 }^{ 4 }\times { 3 }^{ 5 }\times { 3 }^{ 6 }$

Given that:

${ 3 }^{ 1 }\times { 3 }^{ 2 }\times { 3 }^{ 3 }\times { 3 }^{ 4 }\times { 3 }^{ 5 }\times { 3 }^{ 6 }$

$= { 3 }^{ \left( 1+2+3+4+5+6 \right) }$ Using [

$a^{m_1}\times...\times a^{m_n} =a^{m_1+.......m_n}$ ]

$= { 3 }^{ 21 }$

Simplify:
${2}^{2}\times {3}^{3}\times {2}^{4}\times {3}^{5}\times {3}^{6}$

Given that:

${ 2 }^{ 2 }\times { 3 }^{ 3 }\times { 2 }^{ 4 }\times { 3 }^{ 5 }\times { 3 }^{ 6 }$

$= { 2 }^{ \left( 2+4 \right) }\times { 3 }^{ \left( 3+5+6 \right) }$ Using [

$a^{m_1}\times...\times a^{m_n} =a^{m_1+.......m_n}$ ]

$= { 2 }^{ 6 }\times { 3 }^{ 14 }$

Simplify and give reasons:
${ \left[ \left( { 3 }^{ 2 }-{ 2 }^{ 2 } \right) \div \cfrac { 1 }{ 5 } \right] }^{ 2 }$

1375759_566894_ans_864abf70a27e4483b77d31ac34589969.JPG

Rakesh solved some problems of exponents in the following way. Do you agree with the solutions? If not why? Justify your argument.
${ x }^{ -3 }\times { x }^{ -2 }={ x }^{ -6 }$

No, the solved problem is absolutely wrong

Justification:-

$x^{-3}\times x^{-2}=x^{-5}$ (Not

$x^{-6})$

By theory of exponents, we know that

$x^{a}\times x^{b} = x^{a+b}$

$\therefore x^{-3}\times x^{-2}$

$= x^{-3+(-2)}$

$\therefore x^{-3}\times x^{-2}$

$= x^{-5}$

Hence proved.

Find the value of ' $n$ ' in the following:
${ \left( \cfrac { 2 }{ 3 } \right) }^{ 3 }\times { \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }={ \left( \cfrac { 2 }{ 3 } \right) }^{ n-2 }\quad$

we know,

$a^{m}*a^{n}=a^{m+n}$

so,

${ \left( \cfrac { 2 }{ 3 } \right) }^{ 3 }\times { \left( \cfrac { 2 }{ 3 } \right) }^{ 5 }$

$={\left(\cfrac23\right)}^8$

$n-2=8$

$\implies$

$n=10$

If $\log_{2}x = 3$ , then $x =$ ______

We know that, if

$\log _{b}{a}=m$

then,

$a={b}^{m}$

Similarly,

$\log _{2}{x}=3$

So,

$x={2}^{3}$

i.e.

$x=8$ .

Find the value of ' $x$ ' such that
$\cfrac{1}{49}\times {7}^{2x}={7}^{8}$

$\cfrac{1}{49}\times {7}^{2x}={7}^{8}$

$\cfrac{1}{{7}^{2}}\times {7}^{2x}={7}^{8}$

$\left( \because { a }^{ -m }=\cfrac { 1 }{ { a }^{ m } } \right)$

${7}^{-2}\times {7}^{2x}={7}^{8}$

${7}^{2x-2}={7}^{8}$
As bases are equal, Hence

$2x-2=8$

$2x=8+10$

$2x=10$

$x=\cfrac{10}{2}=5$

$\therefore$

$x=5$

Using logarithmic table find the value of the following.
(i) $\log { 23.17 }$
(ii) $\log { 9.321 }$
(iii) $\log { 329.5 }$
(iv) $\log { 0.001364 }$
(v) $\log { 0.9876 }$
(vi) $\log { 6576 }$

$i) \log 23.17 =1.3649 \\ ii) \log 9.321 =0.9694 \\ iii) \log 329.5 =2.5178 \\ iv) \log 0.001364 =-2.8651 \\ v) \log 0.9876 =-0.0054 \\ vi) \log 6576 =3.8179$

Find, by inspection, the characteristics of the logarithms of $21735, 23.8, 350, .035, .2, .87, .875$ .

The characteristic of logarithm of a number

$N > 1$ is one less than the number of digits in the number

$N$

Hence characteristics of the logarithms of

$21735, 23.8, 350$ are

$4,1,2$ respectively

The characteristic of logarithm of a number

$N<1$ is a negative integer with absolute value, one more than the number of zeros to the right of the decimal place

hence the characteristics of

$.035,.2,.87,.875$ are

$-2,-1,-1,-1$ respectively.

Given $\log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980$ , find the value of $\log \sqrt [3]{12}$ .

$\log (\sqrt [ 3 ]{ 12 } )=\dfrac { 1 }{ 3 } \log (12)\\ =\dfrac { 1 }{ 3 } (\log 3+2\log 2)\\ =0.359727\\$

Given $\log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980$ , find the value of $\log 14.4$ .

$\log (14.4)=\log \left(\dfrac { 144 }{ 10 } \right)\\ =\log (12^{ 2 })-1\\ =2(\log 3+2\log 2)-1\\ =1.1583626\\$

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log 84$ .

$\log 84=\log (4\times 21)\\ =2\log 2+\log (21)\\ =2\log 2+\log 3+\log 7\\ =2\times (0.3010300)+0.4771213+0.8450980\\ =1.9242793\\$

Given $\log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980$ , find the value of $\log 4\dfrac{3}{2}$ .

$\log \left(4\dfrac { 2 }{ 3 } \right)=\log \left(\dfrac { 14 }{ 3 } \right)\\ =\log (14)-\log 3\\ =\log 2+\log 7-\log 3\\ =0.3010300+0.8450980-0.4771213\\ =0.6690067$

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log {^{.}128}.$

$\log (.128)=\log 128-\log 100\\ =\log ({ 2 }^{ 7 })-2\\ =7\log 2-2\\ =0.10721$

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log 64$ .

$\log { 64=\log { { 2 }^{ 6 } } } \\ =6\log { 2 }$

But

$\log { 2 } =0.3010300$

$=6\log { 2 } \\ =6(0.3010300)\\ =1.8061800$

Give the position of the first significant figure in the numbers whose logarithms are
$\bar{2} {^{.}7781513}, {^{.}6910815}, \bar{5} {^{.}4871384}.$

$\overline{2}.7781513$ - The characteristic is -2, hence the 1st significant digit will be at the second decimal point

$0.6910815$ - The characteristic is 0, hence the 1st significant digit will be at the units place

$\overline {5}.4871384$ - The characteristic is -5, hence the 1st significant digit will be at the fifth decimal point

How many digits are there in the integral part of the numbers whose logarithms are respectively
$4{^{.}30103}, 1{^{.}4771213}, 3{^{.}69897}, {^{.}56515}?$

The characteristic of logarithm of a number

$N > 1$ is one less than the number of digits in the number

$N$

Characteristics of numbers whose logarithms are $

$4.30103,1.4771213,3.69897,0.56515$ $ are

$4,1,3,0$ respectively.

Hence the number of digits in the integral part of these numbers would be one greater than the characteristic i.e

$5,2,4,1$

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log {^{.}0125}.$

$\log 0.0125=\log \dfrac{125}{10000}=\log {\dfrac1{80}}=\log 1-\log8-\log10=0-3\log2-1=-1.9030900$

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log \sqrt[4]{^{.}0105}.$

$\log \sqrt [ 4 ]{ 0.0105 } =\dfrac { 1 }{ 4 } \left(\log 105-4\right)$

$=\dfrac { 1 }{ 4 } \left(\log \left(7\times 3\times 5\right)-4\right)$

$=\dfrac { 1 }{ 4 } \left(\log 7+\log 3+\log 5-4\right)$

$=\dfrac { 1 }{ 4 } \left(\log 7+\log 3+\log \left(\dfrac{10}{2}\right)-4\right)$

$=\dfrac { 1 }{ 4 } \left(\log 7+\log 3+1-\log 2-4\right)$

$=-0.494702675$

Find the product of $37.203, 3.7203, .037203, 372030$ , having given that
$\log 37.203 = 1.5705780$ , and $\log 1915631 = 6.2823120$ .

$37.203\times 3.7203\times 0.037203\times 372030$

Applying

$\log$ gives

$\log (37.203\times 3.7203\times 0.037203\times 372030)\\ =4\log (37.203)\\ =6.282312\\ =\log 1915631\\ \Rightarrow 37.203\times 3.7203\times 0.037203\times 372030=1915631$

Evaluate : $2^5\times 2^8\div 2^6$

$2^5\times 2^8\div 2^6$

$2^5\times 512\div 64$

$2^5\times 8$

$32\times 8$

$256$ .

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log \sqrt{\dfrac{35}{27}}.$

$\log\left(\sqrt { \dfrac { 35 }{ 27 } } \right)=\dfrac { 1 }{ 2 } \left(\log\left(35\right)-\log\left(27\right)\right)$

$=\dfrac { 1 }{ 2} \left(\log 7\times 5 -\log 3+\log3^{3}\right)$

$=\dfrac { 1 }{ 2} \left(\log 7+\log \left(\dfrac{10}{2}\right)-3\log 3\right)$

$=\dfrac { 1 }{ 2 } \left(\log7+1-\log2-3\log3\right)\\ =0.05635205\\$

Solve the following equations.
$\displaystyle log_2 \, (4^x \, + \, 4) \, - \, log_2 \, (2{x \, + \, 1} \, - \, 3) \, = \, 0.$

we know

$loga-logb=log(a/b)$

therefore

$\log _{ 2 }{ { 4 }^{ x }+4 } -\log _{ 2 }{ (2x+1-3) } =\log _{ 2 }{ \dfrac { { 4 }^{ x }+4 }{ 2x-2 } }$ =0

therefore

$4^x+4=2x-2$

$2^{2x-1}+3=x$

which doesn't not have any solution since exponential function rises rapidly

Write value of $\sqrt [ 3 ]{ 2 } \times \sqrt [ 4 ]{ 2 } \times \sqrt [ 12 ]{ 32 }$

On multiplying , powers of a number get add,

therefore,

$2^\dfrac{1}{3}$

$.2^\dfrac{1}{4}$

$.2^\dfrac{5}{12}$ =

$2^(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{5}{12})$

$2^\dfrac{12}{12}=2^{1}$ ,

$2$

Solve the following equation:
$\displaystyle\, 5^{x - 1} = 10^x\cdot 2^{-x}\cdot 5^{x + 1}$

$5^{x-1}=10^{x}.2^{-x}.5^{x+1}$

$5^{x-1}=5^{x}.5^{x+1}$

$5^{x-1}=5^{2{x}+1}$

$\therefore x=-2$

Rewrite the following equation in the logarithm from :
$5^0 \, = \, 1$

$\textbf{Step 1 : Using definition of logarithm}$

$\boldsymbol{x = a^{b} }$

$\Rightarrow \boldsymbol{\log_{a} x = b}$ ;

$x > 0$

$\, ; a > 0 ,$

$\neq 1$

$\text{Given : } 5^{0} = 1$

$\Rightarrow \log_{5}{1} = 0$

$\textbf{ (Logarithmic Form )}$

Rewrite the following equation in the logarithm from :
$(2\sqrt{2})^{-2/3} \, = \, \dfrac{1}{2}.$

$\textbf{Step 1 : Using definition of logarithm}$

$\boldsymbol{x = a^{b} }$

$\Rightarrow \boldsymbol{\log_{a} x = b}$ ;

$x > 0$

$\, ; a > 0 ,$

$\neq 1$

$Given : (2 \sqrt{2})^{-2/3} = \dfrac{1}{2}$

$\Rightarrow \log_{2\sqrt{2}}\bigg(\dfrac{1}{2}\bigg) = - \dfrac{2}{3}$

Rewrite the following equation in the exponential form.
$log_{5\sqrt{5}} \, 5 \, = \, \dfrac{2}{3}$

$\textbf{Step 1: Using definition of Logarithm}$

$\boldsymbol{\log_{a} x = b}$

$\Rightarrow$

$\boldsymbol{ x = a^{b}}$ ; where

$a > 0 , \neq 1$ and

$x > 0$

$\text{Given : }$

$\log_{5 \sqrt{5}} 5 = \dfrac{2}{3}$

$\Rightarrow (5 \sqrt{5})^\dfrac{2}{3} = 5$

$\textbf{(Exponential form) }$

Rewrite the following equation in the exponential form.
$log_{100} \, 0.1 \, = \, -\dfrac{1}{2}$

$\textbf{Step 1: Using definition of Logarithm}$

$\boldsymbol{\log_{a} x = b}$

$\Rightarrow$

$\boldsymbol{ x = a^{b}}$ ; where

$a > 0 , \neq 1$ and

$x > 0$

$\textbf{Given:}\ \log_{100} 0.1 = \dfrac{-1}{2}$

$\Rightarrow$

$(100)^{-1/2} = 0.1$

Solve $x^{\sqrt{x}} \, = \, (\sqrt {x})^x$

The equation can be re-written as

$x^{\sqrt{x}} \, = \, x^{x/2}$
By trial ,

$x \, = \, 1$ is a solution.
Now equating the exponents , we have

$\sqrt{x} \, = \, \dfrac{1}{2} \, x \, or \, x \, = \, \dfrac{1}{4} \, x^{2} \, or \, x(x \, - \, 4) \, = \, 0$
Since

$x \, = \, 0$ does not satisfy the given equation ,we get

$x \, = \, 4$
Hence the roots of the given equation are

$x \, = \, 1,4$

Find the value of ${(23.17)^{{1 \over {5.76}}}}$ using log table.

$y=(23.17)^{\dfrac{1}{5.76}}$

$\Rightarrow \log y=\cfrac{1}{5.76}\log(23.17)=\cfrac{1.3649}{5.76}=0.23697$

$\Rightarrow y= 1.7257$

Rewrite the following equation in the exponential form.
$\log_3 \, \frac{1}{243} \, = \, -5.$

$\textbf{Step 1: Using definition of Logarithm}$

$\boldsymbol{\log_{a} x = b}$

$\Rightarrow$

$\boldsymbol{ x = a^{b}}$ ; where

$a > 0 , \neq 1$ and

$x > 0$

$Given : \log_3 \, \dfrac{1}{243} = - 5$

$\Rightarrow 3^{-5} = 243$

$\textbf{(Exponential Form) }$

Evaluate : ${2^{{{\log }_3}5}} - {5^{{{\log }_3}2}}$

$2^{\log_{3}5}-5^{\log_{3}2}=5^{\log_{3}2}-5^{\log_{3}2}=0$

${ \left[ { \left( \dfrac { 15 }{ 12 } \right) }^{ 3 } \right] }^{ 4 }$

$\displaystyle =\left [ \left ( \frac{15}{12} \right )^{3} \right ]^{4}$

$\displaystyle =\left [ \left ( \frac{5}{4} \right ) \right ]^{3\times 4}$

$\displaystyle =\frac{5^{12}}{4^{12}}$

If $x=243$ , then find the value of ${x^{\frac{1}{5}}} \times {x^{ - \frac{1}{5}}}$

$\begin{array}{c}{x^{\frac{1}{5}}} \times {x^{ - \frac{1}{5}}} = {x^{\frac{1}{5} - \frac{1}{5}}}\\ = {x^0}\\ = {\left( {243} \right)^0}\\ = 1\end{array}$

Solve: $\log (3x + 2) + \log (3x - 2) = \log5$

$\log (3x + 2) + \log (3x - 2) = \log5$

or,

$\log\{(3x+2)(3x-2)\}=\log 5$ [Using product rule of logarithm]

or,

$9x^2-4=5$

or,

$9x^2=9$

or,

$x^2=1$

or,

$x=\pm 1$ .

Convert the following to logarithmic form:
$5^{2} =25$

$5^2=25$ in logarithmic form is,

$\log_{5}25=2$

If $X^\cfrac{a}{b} =1,$ then find the value of $'a'$ .

${ X }^{ \cfrac { a }{ b } }=1$

$\Rightarrow \dfrac{a}{b}=0$

$\Rightarrow a=0$

Find the value of x for which
${\left( {{3 \over 4}} \right)^6} \times {\left( {{{16} \over 9}} \right)^5} = {\left( {{4 \over 3}} \right)^{x + 2}}$

$(\dfrac{3}{4})^{6}\times (\dfrac{16}{9})^{5}=(\dfrac{4}{3})^{x+2}$

$(\dfrac{4}{3})^{x+2}=(\dfrac{4}{3})^{-6}(\dfrac{4}{3})^{10}$

Comparing powers

$x+2=4\implies x=2$

Simplify:
$\log_{10}5 + 2\log_{10}4$

$\log_{10}5 + 2\log_{10}4$

$=\log_{10}5 +\log_{10}4^2$

$=\log_{10}5 +\log_{10}16$

$=\log_{10}80$ [Using product rule of logarithm]

$=\log_{10}(10\times 8)$

$=\log_{10}10 +\log_{10}8$

$=1 +\log_{10}8$

Find the value of $x$ for which ${ \left( \dfrac { 5 }{ 3 } \right) }^{ -4 }\times { \left( \dfrac { 5 }{ 3 } \right) }^{ -5 }={ \left( \dfrac { 5 }{ 3 } \right) }^{ 3x }$

$\left [ \dfrac{5}{3} \right ]^{(-4+(-5))}=\left [ \dfrac{5}{3} \right ]^{3x}$

$3x=-9$

$x=-3$

$[(64)^{-2}]^{-3} \div [{(-8)^{2}}^{3}]^{2}$

Let us consider

${\left[ {{{\left( {{\rm{64}}} \right)}^{ - {\rm{2}}}}} \right]^{ - {\rm{3}}}} \div {\left[ {{{\left( { - {\rm{8}}} \right)}^{\rm{2}}}^{^{\rm{3}}}} \right]^{\rm{2}}}$

can be written as

$\frac{{{{\left[ {{{\left( {{\rm{64}}} \right)}^{ - {\rm{2}}}}} \right]}^{ - {\rm{3}}}}}}{{{{\left[ {{{\left( { - {\rm{8}}} \right)}^{\rm{2}}}^{^{\rm{3}}}} \right]}^{\rm{2}}}}}$

$=\frac{{{{\left( {{\rm{64}}} \right)}^6}}}{{{{\left( {\rm{8}} \right)}^{{\rm{12}}}}}}$

$(a^{m^n}=a^{mn})$ and power on

$8$ is even so it become positive.

$=\frac{{{{\left( {{\rm{8}} \times {\rm{8}}} \right)}^6}}}{{{{\left( {\rm{8}} \right)}^{{\rm{12}}}}}}$

$=\frac{{{{\rm{8}}^6} \times {{\rm{8}}^6}}}{{{{\left( {\rm{8}} \right)}^{{\rm{12}}}}}}$

$(ab)^m=a^mb^m$

$= \frac{{{{\rm{8}}^{12}}}}{{{{\rm{8}}^{12}}}}$

$(a^m+a^m=a^{m+m})$

$=1$

Simplify :
$\left[ \left( \dfrac { 1 }{ 2 } \right) ^{ -3 }+\left( \dfrac { 1 }{ 3 } \right) ^{ -3 }+\left( \dfrac { 1 }{ 4 } \right) ^{ -3 } \right]$

Let,

$=2^3+3^3+4^3$

$=8+27+64=99$

If $\log_{10}4=0.6021$ and $\log_{10}5=0.6990$ , then find the value of $\log_{10}1600$ .

$\log_{10}1600 =\log_{10}(64\times 25) =\log_{10}(4^3 \times 5^2)$

$=\log_{10}4^3+\log_{10}5^2$

$=3\log_{10}4+2\log_{10}5$

$=3(0.6021)+2(0.6990)$

$=1.8063+1.3980$

$\log_{10}1600=3.2043$

Express $\log_{10}\sqrt[5]{108}$ in terms of $\log_{10}2$ and $\log_{10}3$ .

$\log_{10} \sqrt[5]{108}= \log_{10} \sqrt[5]{2^2\times 3^3}$

$= \log_{10} {(2)^{\dfrac{2}{5}}(3)^{\dfrac{3}{5}}}$

$= \dfrac{2}{5}\log_{10} 2 + \dfrac{3}{5}\log_{10} 3$

Using log tables, calculate $\frac{{24.18 \times 0.004592}}{{0.09588 \times 3.7619}}$

Let

$x=\dfrac{24.18\times\,0.004592}{0.09588\times\,3.7619}$

$\log{x}=\log{\left(\dfrac{24.18\times\,0.004592}{0.09588\times\,3.7619}\right)}$

$=\log{24.18}+\log{0.004592}-\log{0.09588}-\log{3.7619}$

$=1.3834-2.3379+1.01827-0.5754=-0.51163‬$

Find the value of $x$ for which $\left\{ \left( \dfrac { -2 }{ 7 } \right) ^{ 2 } \right\} ^{ x } \times \left( \dfrac { -7 }{ 2 } \right) ^{ 2 } = \dfrac{-8}{343}$ .

${\left\{ {{{\left( {\dfrac{{ - 2}}{7}} \right)}^2}} \right\}^x} \times {\left( {\dfrac{{ - 7}}{2}} \right)^2} = \dfrac{{ - 8}}{{343}}$

$\Rightarrow {\left( {\dfrac{{ - 2}}{7}} \right)^{2x}} \times {\left( {\dfrac{{ - 7}}{2}} \right)^2} = \dfrac{{ - 8}}{{343}}$

$\Rightarrow {\left( {\dfrac{{ - 2}}{7}} \right)^{2x}} \times {\left( {\dfrac{{ - 2}}{7}} \right)^{ - 2}} = {\left( { - \dfrac{2}{7}} \right)^3}$

$\Rightarrow {\left( {\dfrac{{ - 2}}{7}} \right)^{2x - 2}} = {\left( { - \dfrac{2}{7}} \right)^3}$

$\Rightarrow 2x - 2 = 3$

$\Rightarrow 2x = 5$

$\Rightarrow x = \dfrac{5}{2}$

Evaluate ${a^6} \div {a^4}$

Consider the given expression,

$\dfrac {a^6}{a^4}$

$=\dfrac{a^4 \ a^2}{a^4}$

$=a^2$

Hence, the value is

$=a^2$ .

Find $x$ if ${4^{2x}} = \dfrac{1}{{32}}$

(i)

${4^{2x}} = \dfrac{1} {32}$

${\left( {{2^2}} \right)^{2x}} = \dfrac{1} {25}$

${2^{4x}} = {2^{ - 5}}$

$4x = - 5$

$x = \dfrac{- 5}{4}$

Show that $\log_{b^3}a\times \log_{a^3}b\times\log_{a^3}c=\dfrac{1}{27}$ .

Now,

$\log_{b^3}a\times \log_{a^3}b\times\log_{a^3}c$

$=\dfrac{\log_e a}{\log_e b^3}\times \dfrac{\log_eb}{\log_ea^3}\times\dfrac{\log_ec}{\log _ea^3}$ [ Since

$\log_xa=\dfrac{\log_e a}{\log_e x}$ ]

$=\dfrac{\log_e a}{3\log_e b}\times \dfrac{\log_eb}{3\log_ea}\times\dfrac{\log_ec}{3\log _ea}$

$=\dfrac{1}{27}$ .

Simplify
${\left( {\frac{{81}}{{16}}} \right)^{\frac{8}{4}}} \times \left[ {{{\left( {\frac{{25}}{4}} \right)}^{ - \frac{3}{2}}} \div {{\left( {\frac{5}{2}} \right)}^{ - 3}}} \right]$

1315386_1062933_ans_c80056a3d95c434a8497bba0c8796784.jpg

Simplify : ${\left[ {5{{\left( {{8^{\frac{1}{3}}} + {{27}^{\frac{1}{3}}}} \right)}^3}} \right]^{\frac{1}{4}}}$

$= {\left[ {5{{\left( {{8^{1/3}} + {{27}^{1/3}}} \right)}^3}} \right]^{1/4}}$

$= {\left[ {5{{\left( {{2^{3 \times 1/3}} + {3^{3 \times 1/3}}} \right)}^3}} \right]^{1/4}}$

$= {\left[ {5{{\left( {2 + 3} \right)}^3}} \right]^{1/4}}$

$= {\left( {{{5.5}^3}} \right)^{1/4}}$

$= {\left( {{5^4}} \right)^{1/4}}$ f

$= 5$

Solve ${\left( {\sqrt[3]{{\dfrac{2}{3}}}} \right)^{x - 1}}\, = \dfrac{{27}}{8}$

$\displaystyle{\left( {\root 3 \of {\dfrac{2}{3}} } \right)^{x - 1}} = {{27} \over 8}$

$\displaystyle{\left( {{2 \over 3}} \right)^{{{x - 1} \over 3}}} = {\left( {{3 \over 2}} \right)^3}$

$\displaystyle{\left( {{2 \over 3}} \right)^{{{x - 1} \over 3}}} = {\left( {{2 \over 3}} \right)^{ - 3}}$

$\displaystyle{{x - 1} \over 3} = - 3$

$x - 1 = - 9 \Rightarrow x = - 8$

Simplify :
$\dfrac{{{8^{3a}} \times {2^5} \times {2^{2a}}}}{{4 \times {2^{11a}} \times {2^{ - 2a}}}}$

we can write

${8}^{x}={2}^{3x}$

${8}^{3a}={2}^{9a}$

${8}^{3a}\times{2}^{5}\times{2}^{2a}={2}^{9a}\times {2}^{2a+5}={2}^{11a+5}$

${4}={2}^{2}$

$4\times{2}^{11a}\times{2}^{-2a}={2}^{2}\times{2}^{9a}={2}^{9a+2}$

$\dfrac{{2}^{11a+5}}{{2}^{9a+2}}={2}^{2a+3}$

Find the value of $log_7343$ .

Let factorise 343. we get

So,

$343 = 7 \times 7 \times 7$

$343 = 7^{3}$

So,

$log_{7}343 = log_{7}7^{3}$

now, as we know,

$log_{a}b^{n} = n log a^{b}$

& If

$log_{a}^{a}$ is given, then

$log_{a}^{a} = 1$

So,

$log_{7}343 = log_{7}7^{3} = 3 log_{7}7$

therefore,

$\boxed{log_{7}343 = 3}$

1216193_1068989_ans_3dd151fd9f4a479599bd8b51bdf0b5a6.jpg

Find the value of ${ \log }_{ 10 }\dfrac { 76 }{ 3.8 }$

$\log_{10}\dfrac{76}{3.8}$

$\Rightarrow \log_{10}20$

$\Rightarrow \log_{10}(10\times 2)$

$\Rightarrow \log_{10}10+\log_{10}2$

$=1+0.3010$

$\Rightarrow \log_{10}10+\log_{10}2$

$=1.3010$

Evaluate:-
${\left( {\frac{2}{7}} \right)^2}\, \times \,{\left( {\frac{7}{2}} \right)^{ - 3}}\, \div \,\,{\left\{ {{{\left( {\frac{7}{5}} \right)}^{ - 2}}} \right\}^{ - 4}}$

${\left( {\dfrac{2}{7}} \right)^2} \times {\left( {\dfrac{7}{2}} \right)^{ - 3}} \div {\left( {{{\left( {\dfrac{7}{5}} \right)}^{ - 2}}} \right)^{ - 4}}$

${\left( {\dfrac{2}{7}} \right)^2} \times {\left( {\dfrac{2}{7}} \right)^3} \div {\left( {\dfrac{7}{5}} \right)^{\left( { - 8} \right)}}$

$= {\left( {\dfrac{2}{7}} \right)^5} \div {\left( {\dfrac{5}{7}} \right)^8}$

$= {\left( {\dfrac{2}{7}} \right)^5} \times {\left( {\dfrac{7}{8}} \right)^8}$

$= \dfrac{{{2^5}}}{{{7^5}}} \times \dfrac{{{7^8}}}{{{8^8}}}$

$= \dfrac{{32}}{{8 \times {8^7}}} \times {7^{8 - 5}}$

$= \dfrac{4}{{8 \times {8^5}}} \times 7$

$= \dfrac{{{7^3}}}{{{8^6} \times 2}}$

Solve: $(64)^{\dfrac{2}{3}} + 9^{\dfrac{3}{2}}$ .

Now,

$(64)^{\dfrac{2}{3}} + 9^{\dfrac{3}{2}}$

$=(4^3)^{\dfrac{2}{3}} + (3^2)^{\dfrac{3}{2}}$

$=4^2+3^3$

$=16+27$

$=43$ .

${\left( {\frac{4}{5}} \right)^2}\, \times \,{5^4}\, \times \,{\left( {\frac{2}{5}} \right)^{ - 2}}\, \div \,{\left( {\frac{5}{2}} \right)^{ - 3}}$

${\left( {\dfrac{4}{5}} \right)^2} \times {5^4} \times {\left( {\dfrac{2}{5}} \right)^{ - 2}} \div {\left( {\dfrac{5}{2}} \right)^{ - 3}}$

$\dfrac{{{4^2}}}{{{5^2}}} \times {5^4} \times \dfrac{{{5^2}}}{{{2^2}}} \div {\left( {\dfrac{2}{5}} \right)^3}$

$\dfrac{{4 \times 4 \times {5^4}}}{{2 \times 2}} \times \dfrac{{{{\left( 5 \right)}^3}}}{{{2^3}}}$

$\dfrac{{4 \times {5^7}}}{{2 \times 2 \times 2}} = \dfrac{{{5^7}}}{2}$

Using laws of exponents, simplify and write the answer in exponential form:
(i) $3^2 \times 3^4 \times 3^8$ (ii) $\dfrac{6^{15}}{6^{10}}$

IIt is known that

$a^m\times a^n=a^{m+n}$ and

$\dfrac{a^m}{a^n}=a^{m-n}$

We have,

(i)

$3^2 \times 3^4 \times 3^8=3^{2+4+8}=3^{14}$ .[ Using product rule]

(ii)

$\dfrac{6^{15}}{6^{10}}=6^{15-10}=6^5$ . [ Using quotient rule]

Solve
$\log_{\frac{1}{3}}(x^2 + 8) = -2$

$\log _{ 3 }{ ({ x }^{ 2 }+8) } =2$

${ x }^{ 2 }+8={ 3 }^{ 2 }=9$

${ x }^{ 2 }=1$

$x=\pm 1$

Solution of $3^{3x-5}=\frac{1}{9^x}$ is

$3^{3x-5}=(\frac{1}{9^x})\\3^{3x-5}=3^{-2x}\\\therefore 3x-5=-2x\\5x=5\\\therefore x=1$

If $log_2 x=a$ and $log_5y=a$ , write $100^{2a-1}$ in terms of x and y.

$log-2x=a\\\therefore\>x=2^a\\and\>logg_5y=a\\\>then\>y=5^a\\then\>xy=2^a5^a\\xy=10^a\\\therefore\>100^{2a-1}=(\frac{100^{2a}}{100})\\=(\frac{10^{4a}}{100})\\=(\frac{(10^a)^4}{100})\\=(\frac{(xy)^4}{100})$

Determine the value of ${3^2} -\{ {\log}3\}^6$

Given that

$={{3}^{2}}-{{\log }^{6}}3$

$={{3}^{2}}-{{\left( \log 3 \right)}^{6}}$

We know that

$\log 3=0.4771$

$\therefore \,{{3}^{2}}-{{\left( 0.4771 \right)}^{6}}$

$=9-0.117938744$

$=8.98820613$

$=8.988$

hence this is the answer

Simplify:
$\left[(64)^{-3}\times (81)^{-\frac{9}{4}}\right]^{-\frac{1}{9}}$

Now,

$\left[(64)^{-3}\times (81)^{-\frac{9}{4}}\right]^{-\frac{1}{9}}$

$=\left[(4^3)^{-3}\times (3^4)^{-\frac{9}{4}}\right]^{-\frac{1}{9}}$

$=\left[(4)^{-9}\times (3)^{-{9}}\right]^{-\frac{1}{9}}$

$=\left[(4)\times (3)\right]^{-\frac{1}{9}\times (-9)}$

$=4\times 3$

$=12$ .

Simplify:
$\dfrac{{3 \times {7^2} \times {{11}^8}}}{{21 \times {{11}^3}}}$

$\dfrac{3\times {7}^{2}\times{11}^{8}}{21\times {11}^{3}}$

$=\dfrac{3\times {7}^{2}\times{11}^{8}}{3\times 7\times {11}^{3}}$

$={3}^{1-1}\times {7}^{2-1}\times{11}^{8-3}$

$={3}^{0}\times {7}^{1}\times{11}^{5}$

$=7 \times{11}^{5}$

$=7\times 1,61,051=11,27,357$

Simplify: $a^3 \times a^3 \times 5a^4$ .

It is known that,

$a^m\times a^n=a^{m+n}$

Now,

$a^3 \times a^3 \times 5a^4$

$=5a^{3+3+4}$

$=5a^{10}$ .

${5^{x - 3}}\,\,\, \times \,{3^{2x - 8}} = 225$

We have,

${{5}^{x-3}}\times {{3}^{2x-8}}=225$

$\dfrac{{{5}^{x}}}{{{5}^{3}}}\times \dfrac{{{3}^{2x}}}{{{3}^{8}}}=225$

${{5}^{x}}\times {{3}^{2x}}=25\times 9\times {{5}^{3}}\times {{3}^{8}}$

${{5}^{x}}\times {{3}^{2x}}={{5}^{2}}\times {{3}^{2}}\times {{5}^{3}}\times {{3}^{8}}$

${{5}^{x}}\times {{3}^{2x}}={{5}^{5}}\times {{3}^{10}}$

${{5}^{x}}\times {{3}^{2x}}={{5}^{5}}\times {{3}^{2\times 5}}$

So, $x = 5$

Hence, the value of $x$ is $5$ .

Solve ${\left( { - \frac{3}{5}} \right)^{ - 3}}$

We have,

${{\left( -\dfrac{3}{5} \right)}^{-3}}$

$={{\left( -\dfrac{5}{3} \right)}^{3}}$

$=-\dfrac{125}{27}$

Hence, this is the answer.

Simplify $\log c\sqrt c$

Given expression

$\log { c\sqrt { c } }$

$=$

$\log { c({ c }^{ \dfrac { 1 }{ 2 } } } )$

$=$

$\log { ({ c }^{ 1+\dfrac { 1 }{ 2 } } } )$

$=$

$\log { ({ c }^{ \dfrac { 3 }{ 2 } } } )$

$=$

$\dfrac { 3 }{ 2 } \log { { c } }$

Find the zeroes of the polynomial $p\left( x \right) = x - \log _2{16}$ .

Given ,

$\log_{2}16=\log_{2}2^4=4$

$p(x)=x-\log_{2}16$

$=>x-4=0$

$x=4$

Solve: $\dfrac{{{3^5}\times{{10}^5}\times25}}{{{5^7}\times{6^5}}}$

Given that:

$\Rightarrow\dfrac{3^5\times10^5\times25}{5^7\times6^5}$

$\Rightarrow\dfrac{3^5\times(5\times 2)^5\times 5^2}{5^5\times(3\times 2)^5}$

$\Rightarrow\dfrac{3^5\times\times 5^5\times 2^5\times 5^2}{5^5\times 3^5\times 2^5}$

$=1$

Hence, this is the answer.

Given $log_{10}x=2a$ and $log_{10}y=\dfrac{b}{2}$ . Write $10^a$ in terms of x.

1327393_1083103_ans_b7afcfb61b9245b2a01491e29fa8cfd3.jpg

Given $log_{10}x=2a$ and $log_{10}y=\dfrac{b}{2}$ , write $10^{2b+1}$ in terms of y.

1327388_1083104_ans_7a5f5b9d64784db8a8c2413861b435a5.jpg

$log\dfrac{75}{16}-2log\dfrac{5}{9}+log\dfrac{32}{243}=log 2$ .

1324445_1086591_ans_f2c9247efd9b4282a50e10559c0a2518.jpg

Find $\log _{ 7 }{ 1 }$

$For\ any\ a\in\ (0, 1)\ \cup\ (1, \infty),\ log_a{1} = 0.$

$Hence,\ log_7{1} = 0$

Express the following in exponential form.
$\dfrac{5}{6}\times \dfrac{5}{6}\rightarrow \left(\dfrac{5}{6}\right)^2$ .

$\Bigg(\cfrac{5}{6} \Bigg)^2 = \cfrac{5^2}{2^2 \times 3^2}$

Solve:
$\displaystyle {\left\{ {{{\left( {\frac{1}{3}} \right)}^{ - 1}} - {{\left( {\frac{1}{4}} \right)}^{ - 1}}} \right\}^{ - 1}}$

Consider the given expression.

${{\left\{ {{\left( \dfrac{1}{3} \right)}^{-1}}-{{\left( \dfrac{1}{4} \right)}^{-1}} \right\}}^{-1}}$

$={{\left\{ 3-4 \right\}}^{-1}}$

$={{\left\{ -1 \right\}}^{-1}}$

$=\dfrac{1}{-1}$

$=-1$

Hence, this is the answer.

Simplify : $4\sqrt{16} - 6\sqrt[3]{343} + 18 \sqrt[5]{243} - \sqrt{196}$

$\sqrt[4]{16}-6\sqrt[3]{343}+18\sqrt[5]{243}-\sqrt{196}$

$=\sqrt[4]{2\times 2\times 2\times 2}-6\sqrt[3]{7\times 7\times 7}+18\sqrt[5]{3\times 3\times 3\times 3\times 3}-\sqrt{2\times 2\times 7\times 7}$

$=\sqrt[4]{(2)^{4}}-6\sqrt[3]{7^{3}}+18\sqrt[5]{3^{5}}-2\times 7$

$=2-6(7)+18(3)-14$

$=2-42+54-14$

$=0$

value of ln10?

$ln 10$

$= 2.302$

Solve ${(3.968)^{\frac{3}{2}}}$

Given that,

$(3.968)^\frac{3}{2}=\sqrt{(3.968)^3}$

$=3.968(\sqrt{3.968})$

$=3.968\times 1.99$

$=7.9041$

Hence, this is the answer.

Solve :
$\log M = \log {\left( {0.9} \right)^{20}}$

$\\logM=log(0.9)^{20}$

$\\M=(0.9)^{20}$

$\\M=0.121(approx)$

If ${\left( {2.381} \right)^x} = {\left( {0.2381} \right)^y} = {10^z}$ , then find the value of $\frac{1}{y} + \frac{1}{z} - \frac{1}{x}$

Let,

$(2.381)^x=(0.2381)^y=10^z=k$

$\implies 2.381=k^{\frac{1}{x}} \ , \ 0.2381=k^{\frac{1}{y}} \ , \ 10=k^{\frac{1}{z}}$

Now,

$10=\dfrac{2.381}{0.2381}$

$k^{\frac{1}{z}}=\dfrac{k^{\frac{1}{x}}}{\frac{1}{y}}$

$k^{\frac{1}{z}}=k^{\frac{1}{x} - \frac{1}{y}}$ ------ Quotient law of exponents

$\dfrac{1}{z}=\dfrac{1}{x} - \dfrac{1}{y}$ ------ Base is same, powers can be equated

$\therefore \dfrac{1}{z} - \dfrac{1}{x} + \dfrac{1}{y}=0$

What will be the value of $log_2 \ (log_3 \ 81)$ ?

We have,

${{\log }_{2}}\left( {{\log }_{3}}81 \right)$

$={{\log }_{2}}\left( {{\log }_{3}}{{3}^{4}} \right)$

$={{\log }_{2}}\left( 4{{\log }_{3}}3 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because {{\log }_{a}}{{x}^{n}}=n{{\log }_{a}}x \right]$

$={{\log }_{2}}4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because {{\log }_{a}}a=1 \right]$

$={{\log }_{2}}{{2}^{2}}$

$=2{{\log }_{2}}2$

$=2\times 1$

$=2$

Hence, this is the answer.

Find the value of $\log_{2}{32}$ .

From question,

$log_{2} {32}$

$log_{2}2^5$

$5log_2$

$2$

$5\times 1$

$5$

$log_{\dfrac12}8=?$

$\log_{{1}/{2}}{8}$

$= \log_{{2}^{-1}}{8}$

$= - \log_{2}{\left( {2}^{3} \right)} \; \left( \because \log_{{a}^{b}}{c} = \dfrac{1}{b} \log_{a}{c} \right)$

$= -3 \log_{2}{2} \; \left( \because \log_{a}{\left( {b}^{c} \right)} = c \log_{a}{b} \right)$

$= -3 \; \left( \because \log_{a}{a} = 1 \right)$

Hence the correct answer is

$-3$ .

If $log_{10}8=0.90$ find the value of :
(i) $log_{10}4$
(ii) $log\sqrt{32}$
(iii) $log \ 0.125$

Given:-

$\log_{10}{8} = 0.90$

To find:-

(i)

$\log_{10}{4}$

(ii)

$\log_{10}{\sqrt{32}}$

(iii)

$\log_{10}{0.125}$

$\log_{10}{8} = 0.90 \; \left( \text{Given} \right)$

$\log_{10}{\left( {2}^{3} \right)} = 0.9$

$\Rightarrow 3 \log_{10}{2} = 0.9$

$\Rightarrow \log_{10}{2} = \dfrac{0.9}{3} = 0.3$

(i)

$\log_{10}{4}$

$= \log_{10}{\left( {2}^{2} \right)}$

$= 2 \log_{10}{2}$

$= 2 \times 0.3 = 0.6$

(ii)

$\log_{10}{\sqrt{32}}$

$= \log_{10}{\left( {\left( 32 \right)}^{\dfrac{1}{2}} \right)}$

$= \dfrac{1}{2} \log_{10}{\left( {2}^{5} \right)}$

$= \dfrac{5}{2} \log_{10}{2}$

$= \dfrac{5}{2} \times 0.3$

$= 0.75$

(iii)

$\log_{10}{0.125}$

$= \log_{10}{\left( {\left( 0.5 \right)}^{3} \right)}$

$= 3 \log_{10}{\left( \dfrac{5}{10} \right)}$

$= 3 \left( \log_{10}{5} - \log_{10}{10} \right)$

$= 3 \left( \log_{10}{5} - \log_{10}{\left( 2 \times 5 \right)} \right)$

$= 3 \left( \log_{10}{5} - \left( \log_{10}{2} + \log_{10}{5} \right) \right)$

$= 3 \left( - \log_{10}{2} \right)$

$= 3 \times \left( - 0.3 \right)$

$= -0.9$

Compute the following
$7^{\log_{3}5}+5^{\log_{5}7}-5^{\log_{3}7}-7^{\log_{5}3}$

$7^{log_3^5}+5^{log_5^7}-5^{log_3^7}-7^{log_5^3}$

$=5^{log_3^7}+7^{log_5^5}-5^{log_3^7}-7^{log_5^3}$

$=7-7^{log_5^3}$ .

1191742_1157731_ans_ca1f72c38cdf44a19227f971e09b941c.jpg

Using laws of exponents, simplify and write the answer in exponential form:
(i) ${7}^{x}\times {7}^{2}$
(ii) ${2}^{5}\times {5}^{5}$
(iii) ${a}^{4}\times {b}^{4}$

(i)

$7^{x}\times 7^{2}=7^{x+2}$

(ii)

$2^{5}\times 5^{5}=(2\times 5)^{5}=10^{5}$

(iii)

$a^{4}\times b^{4}=(a\times b)^{4}=(ab)^{4}$

Using laws of exponents, simplify and write the answer in exponential form:
(i) ${3}^{2}\times {3}^{4}\times {3}^{8}$
(ii) ${6}^{15}\div {6}^{10}$
(iii) ${a}^{3}\times {a}^{2}$

(i)

$3^{2}\times 3^{4}\times 3^{8}$ =

$3^{2+4+8}=3^{14}$

(ii)

$6^{15}\div 6^{10}=6^{15-10}=6^{5}$

(iii)

$a^{3}\times a^{2}=a^{3+2}=a^{5}$

Evaluate the following
$\dfrac {\left(\dfrac {12}{13}\right)^{5}\times \left(\dfrac {-1}{3}\right)}{\dfrac {1}{81}\times \left(\dfrac {12}{13}\right)^{3}}$

Evaluating given problem using law of eponents,

$\implies \dfrac{(12)^5\times(-1)\times 81\times(13)^5}{(13)^3\times 3\times(12)^3}$

$= \dfrac{(12)^2\times(-1)\times 81\times(13)^2}{3}$

$= 144\times(-1)\times 27\times169$

$= -657072$

Find the value of
$log5.4$

$\log5.4=\log5+\log4=0.698+0.602=1.300$

simplify :
${\left( {{3^4}} \right)^3}$

${\left({3}^{4}\right)}^{3}={3}^{12}=5,31,441$ using

${\left({a}^{m}\right)}^{n}={a}^{mn}$

Solve: $\dfrac{3^2 \times 3^2 \times 2^2}{3^2 \times 6}$

$\dfrac{3^2 \times 3^2 \times 2^2}{3^2 \times 6} = \dfrac{3^2 \times 3^2 \times 2^2}{3^2 \times 3 \times 2} = 3 \times 2 = 6$

Simplify the following using laws of exponents.
${9^2} \times {9^{18}} \times {9^{10}}$

We have,

$9^2\times9^{18}\times9^{10}$

$=9^{2+18+10}$

$=9^{30}$

Hence, this is the answer.

Simplify the following using laws of exponents.
$({3^2}) \times {({3^2})^4}$

Consider the following question.

$\left( {{3}^{2}} \right)\times {{\left( {{3}^{2}} \right)}^{4}}={{3}^{2}}\times {{3}^{8}} \\$

$={{3}^{10}} \\$

Hence, this is the required answer.

Apply laws of exponents and simplify.
$(3^{0}\times2^{5})+5^{0}$

$(3^o\times 2^5)+5^o$

Let

$a=(3^o\times 2^5)+5^o$

$=(1\times 2^5)+1$

$=32+1$

$a=33$

$\therefore (3^o\times 2^5)+5^o=3$

simplify:
$\left( {\frac{{{2^{20}}}}{{{2^{15}}}}} \right) \times {2^3}$

$\dfrac{{2}^{20}}{{2}^{15}}\times{2}^{3}$

$={2}^{20+3-15}$

$={2}^{8}$

$=256$

Simplify the following using laws of exponents.
${2^{10}} \times {2^4}$

By using the property ${a^m} \times {a^n} = {a^{m + n}}$ on the given expression,

${{2}^{10}}\times {{2}^{4}}={{2}^{10+4}}$

$= {2^{14}}$

Solve : $\dfrac{(3^5)^2\times 7^3}{(3^3)^3 \times 7^2}$

Laws of indices:

$(a^m)^n=a^{mn}$

$\dfrac{a^m}{a^n}=a^{m-n}$
----------------------------------------------------------------------

$\dfrac{(3^5)^2\times 7^3}{(3^3)^3 \times 7^2}$

$=\dfrac{3^{10}\times 7^3}{3^9 \times 7^2}$

$=3^{10-9}\times7^{3-2}$

$=3\times7$

$={21}$

If $\log _{ 10 }{ 2 } =0.3010$ , then $\log _{ 10 }{ 50 }$ is

Given

$\log_{10}2=0.3010$

$\log_{10}\left(\dfrac{100}{50}\right)=0.3010$

$\log_{10}100-\log_{10}50=0.3010$

$2-\log_{10}50=0.3010$

$\log_{10}50=2-0.3010$

$\log_{10}50=1.699$

Find the value of $\log_{\sqrt{3}}81$ .

$log^{(81)}_{(\sqrt{3})}$

$=log^{3^4}_{(\sqrt{3})}$

$=log^{(\sqrt{3})^8}_{\sqrt{3}}$

$=8 log^{\sqrt{3}}_{\sqrt{3}}$

$=8[\because log^a_a=1$ for

$a > 1]$ .

1180397_1195806_ans_d1ab5df126984b159c03a19fab7b9ba2.jpg

Using the log table find the value of $1.234$ .

$log_{10}1.234$

As number is already in the scientific form, look first two digits

$(1.2)$ in first column in log table, then the entry corresponding to the

$3^{rd}$ digit

$(3)$ in the row of

$1.2$ , here we have the entry as

$0.899$ add the mean difference corresponding to the forth digit

$(4)$ , Here, we have

$14$ , and add it to

$0.899\Rightarrow 0.899+0.0014=0.0913$

$\Rightarrow log_{10}1.234=0.0913=0.0913$ .

1114495_1197002_ans_0bcfe64986d440c3a7f50e019e6c7f78.jpg

Solve $\left( 3 ^ { - 7 } \div 3 ^ { - 10 } \right) \times 3 ^ { - 5 }$

$=\dfrac{{3}^{-7}}{{3}^{-10}}\times{3}^{-5}$

$={3}^{-7+10-5}$

$={3}^{-12+10}$

$={3}^{-2}$

$=\dfrac{1}{{3}^{2}}=\dfrac{1}{9}$

Find the value of $[\{(a^3)^{-3}\}^0]^{100}$ .

${ \left[ { \left\{ { \left( { a }^{ 3 } \right) }^{ -3 } \right\} }^{ 0 } \right] }^{ 100 }\\ ={ \left[ { \left\{ { a }^{ -9 } \right\} }^{ 0 } \right] }^{ 100 }\quad \left[ \because { \left( { a }^{ m } \right) }^{ n }={ a }^{ mn } \right] \\ ={ \left[ { a }^{ 0 } \right] }^{ 100 }\\ ={ 1 }^{ 100 }\\ =1.$

If ${\log _{10}}8 = 0.90$
find $\log 0.125$

Given :

$log_{10}8 = 0.90.$

$\frac{1}{8}\times 8 = 1$

$\Rightarrow (0.25) \times 8 = 1$

apply log on both sides.

$\Rightarrow log_{10}(0.125)+log_{10}(8) = log_{10}(1)$

$\Rightarrow log_{10}(0.125) = 0-log_{10}(8)$

$\Rightarrow log_{10}(0.125) = -(0.9)$

1108490_1189400_ans_d042cc25db6d4370ad68600468fd7152.jpeg

If ${\log _{10}}8 = 0.90;$ find the value of :
a) ${\log _{10}}4$
b) $\log \sqrt {32}$
c) $\log 0.125$

$\log _{ 10 }{ 8 } =0.9 = 3\log _{ 10 }{ 2 } =0.9$

$\boxed {=\log _{ 10 }{ 2 }=0.3}$

$\left( i \right) \log _{ 10 }{ 4 } = \log _{ 10 }{ 2^2 } = 2\log _{ 10 }{ 2 } = 2 \times 0.3 = 0.6$

$\left (ii \right) \log_{10}{\sqrt {32}} = \dfrac {1}{2} \log_{10}{32}= \dfrac {1}{2} \log_{10}{2^5}= \dfrac {5}{2} \log_{10}{2} =\dfrac {5}{2} \times 0.3 = 0.75$

$\left(iii \right) \log_{10}{0.125}=\log_{10}{125\times 10^{-3}}=\log_{10}{5^{3}}+\log_{10}{10^{-3}}=3\log_{10}{5}-3$

$=-3+3\log _{ 10 }{ \left( \dfrac { 10 }{ 2 } \right) } =-3+3\left[\log_{10}{10}-\log_{10}{2}\right] = -3\log_{10}{2}=-0.9$

$a^{m}.a^{n}=$

$a^{m}.a^{n}=a^{m+n}$

Find the value of the following.
$4^{4}$

$4^{4} = (4\times 4)\times (4\times 4) = 16\times 16 = 256$

Solve the exponent

${17^2} \cdot {17^{ - 5}}$

$\begin{array}{l} { 17^{ 2 } }\cdot { 17^{ -5 } } \\ ={ 17^{ 2+\left( { -5 } \right) } } \\ ={ 17^{ 2-5 } } \\ ={ 17^{ -3 } } \\ =\frac { 1 }{ { { { 17 }^{ 3 } } } } \\ =\frac { 1 }{ { 4913 } } \end{array}$

By what number should $( - 6 ) ^ { - 2 }$ be multiplied so that the product would be equal to $( 9 ) - 1 .$ ?

$(-6)^{-2}$ so that the product will become

$(9)-1$ .

$\rightarrow (-6)^{-2}$ can be written as

$\dfrac{1}{(-6)^2}$

$=\dfrac{1}{36}$

Let the required number be

$x$

According to the given condition,

$\dfrac{x}{36}=(9)-1$

$x=36\times 8$

$x=288$

$(-\dfrac{3}{4})^{11}\div [ (-\dfrac{3}{4})^{3}\times (-\dfrac{3}{4}^{6}) ]$

$=\left(-\dfrac{3}{4}\right)^{11}\div \left[\left(-\dfrac{3}{4}\right)^3\times \left(-\dfrac{3}{4}\right)^6\right]$

$=\left(-\dfrac{3}{4}\right)^{11}\div \left(-\dfrac{3}{4}\right)^{3+6}$

$=\left(-\dfrac{3}{4}\right)^{11}\div \left(-\dfrac{3}{4}\right)^{9}$

$=\left(-\dfrac{3}{4}\right)^{11-9}=\left(-\dfrac{3}{4}\right)^2$

$=\dfrac{(-3)\times (-3)}{4\times 4}=\dfrac{9}{16}$ .

1127975_1242406_ans_0fb3eec854bb43c9a43212cac9899b82.jpg

Given: $\log{2}=0.3010$ and $\log{3}=0.4771$ , find the value of $\log{12}$ .

Given

$\log2=0.3010$ and

$\log3=0.4771$

$\log12=\log(4\times3)$

$=\log4+\log3$

$=\log2^2+\log3$

$=2\log2+\log3$

Substituting the values

$=2\times0.3010+0.4771$

$=0.602+0.4771$

$=1.0791$

Find the value of y if : $(100)^2\times(10)^5=(1000)^y$

Consider the given equation,

${{\left( 100 \right)}^{2}}\times {{\left( 10 \right)}^{5}}={{\left( 100 \right)}^{y}}$

${{\left( {{10}^{2}} \right)}^{2}}\times {{\left( 10 \right)}^{5}}={{\left( 10 \right)}^{y}}$

$\left( {{10}^{4}} \right)\times {{\left( 10 \right)}^{5}}={{\left( 10 \right)}^{y}}$

$\left( {{10}^{4+5}} \right)={{\left( 10 \right)}^{y}}$

${{10}^{9}}={{10}^{y}}$

$y=9$

Hence, this is the answer.

Find the value of $m$ if :
$\left( \dfrac { 2 }{ 9 } \right)^3 \times \left( \dfrac { 2 }{ 9 } \right)^{-6} =\left( \dfrac { 2 }{ 9 } \right)^{2m-1}$

We have,

$\left(\dfrac{2}{9}\right)^3\times \left(\dfrac{2}{9}\right)^{-6}=\left(\dfrac{2}{9}\right)^{2m-1}$

$\left(\dfrac{2}{9}\right)^{3-6}=\left(\dfrac{2}{9}\right)^{2m-1}$

$\left(\dfrac{2}{9}\right)^{-3}=\left(\dfrac{2}{9}\right)^{2m-1}$

On comparing both power, we get

$2m-1=-3$

$2m=-2$

$m=-1$

Simplify the following using laws of expressions
$(2x)^{4}\div (2x)^{2}$

Given that,

$\begin{array}{l} { \left( { 2x } \right) ^{ 4 } }\div { \left( { 2x } \right) ^{ 2 } } \\ =\frac { { { { \left( { 2x } \right) }^{ 4 } } } }{ { { { \left( { 2x } \right) }^{ 2 } } } } \end{array}$

We know that,

$\begin{array}{l} \frac { { { x^{ a } } } }{ { { x^{ b } } } } ={ x^{ a-b } } \\ \frac { { { { \left( { 2x } \right) }^{ 4 } } } }{ { { { \left( { 2x } \right) }^{ 2 } } } } ={ \left( { 2x } \right) ^{ 4-2 } }={ \left( { 2x } \right) ^{ 2 } } \\ ={ \left( { 2x } \right) ^{ 2 } } \end{array}$

We know that,

$\begin{array}{l} { \left( { a.b } \right) ^{ n } }={ a^{ n } }{ b^{ n } } \\ ={ 2^{ 2 } }{ x^{ 2 } } \\ { 2^{ 4 } }=4 \\ =4{ x^{ 2 } } \end{array}$

Then ,

We get

$4{x^2}.$

Determine the value of the following
$\log_{7}{1}$

$log_{7}1= \frac{log1}{log7} = 0$
1122000_1248001_ans_abf1c4121f9a42fbba55ec59e9ad0420.jpg

Express as a power of $3$ in $729$ and $343$ .

We have,

Power of $3$ in $729$ .

${{9}^{3}}=729$

Power of $3$ in 343.

${{7}^{3}}=343$

Hence, this is the answer.

Find the value of $(x^{3}\times x^{7})\div x^{12}$ for x = (-2).

$(x^3+x^7)\div x^{12}$ for

$x=-2$

$x^{10}\div x^{12}=\dfrac{x^{10}}{x^{12}}=\dfrac{1}{x^2}$

For

$x=-2$

$=\dfrac{1}{(-2)^2}=\dfrac{1}{4}$ .

1127977_1242482_ans_6df4ecd9c0264a85a37279b539387b12.jpg

Find the value $x$ if ${2^4}*{2^5} = {({2^3})^x}$

$2^{4}\times 2^{5}=(2^{3})^{x}$

but

$m^{x}\times m^{y}=m^{x+y}$

$\therefore 2^{4+5}=2^{3x}$

$\Rightarrow 2^{9}=2^{3x} {\Rightarrow }3x=9\Rightarrow x=3$

1212377_1298694_ans_50a30c2523aa432ab2f723d2d730a139.JPG

Determine the value of the following
$\log_{10}{0.01}$

$log_{10}0.01 = log_{10}10^{-2}$

$= -2 log^{10}10$

$= -2$

1122018_1248004_ans_22f2f84f510b4bd28289e4499f5ec808.jpg

If $log_{10}0.001 =x$ , then find $x$ .

1339256_1258792_ans_6bbb6a82d0ad4b19ba9657bf6b365cd3.jpg

If $x ^ { 2 } + y ^ { 2 } = 47 x y$ then show that $\log \left( \dfrac { x + y } { 7 } \right) = \dfrac { 1 } { 2 } ( \log x + \log y ).$

Given that,

${x^2} + {y^2} = 47xy$

Add

$2xy$ both sides,

$\begin{array}{l} { x^{ 2 } }+{ y^{ 2 } }+2xy=47xy+2xy \\ { \left( { x+y } \right) ^{ 2 } }=49xy \end{array}$

Square root both sides,

$x+y=7\sqrt { xy }$

Substitute the above equation in,

$\begin{array}{l} \log \left( { \frac { { x+y } }{ 7 } } \right) =\frac { 1 }{ 2 } \left( { \log x+\log y } \right) \\ \log \left( { \frac { { 7\sqrt { xy } } }{ 7 } } \right) =\frac { 1 }{ 2 } \left( { \log x+\log y } \right) \\ \log \left( { \frac { { 7{ { \left( { xy } \right) }^{ \frac { 1 }{ 2 } } } } }{ 7 } } \right) =\frac { 1 }{ 2 } \left( { \log x+\log y } \right) \\ \log \left( { { { \left( { xy } \right) }^{ \frac { 1 }{ 2 } } } } \right) =\frac { 1 }{ 2 } \left( { \log x+\log y } \right) \\ \frac { 1 }{ 2 } \left( { \log \left( { xy } \right) } \right) =\frac { 1 }{ 2 } \left( { \log x+\log y } \right) \\ \frac { 1 }{ 2 } \left( { \log x+\log y } \right) =\frac { 1 }{ 2 } \left( { \log x+\log y } \right) \end{array}$

LHS=RHS

Find the value of $log_{2\sqrt{3}}1728$

Let

$y =log _{2\sqrt{3}}(1728)$ .

we know that

$1728= 64\times 24$

$(2)^{6}\times (\sqrt{3})^{6}=(2\sqrt{3})^{6}$

then

$y = log _{2\sqrt{3}}$

$(2\sqrt{3})^6$

$y=6 log_{2\sqrt{3}}$

$(2\sqrt{3})$

$[log _{x }x=1]$

$[y=6]$ .

1180783_1353625_ans_1da02aa661044d778173356cc867e9b9.JPG

Solve : $\left( 6 ^ { - 1 } - 8 ^ { - 1 } \right) ^ { - 1 } + \left( 2 ^ { - 1 } - 3 ^ { - 1 } \right) ^ { - 1 }$

$\begin{array}{l} ={ \left( { \dfrac { 1 }{ 6 } -\dfrac { 1 }{ 8 } } \right) ^{ -1 } }+{ \left( { \dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 } } \right) ^{ -1 } } \\ =\left( { \dfrac { 1 }{ { \left( { \dfrac { 1 }{ 6 } -\dfrac { 1 }{ 8 } } \right) } } } \right) +\left( { \dfrac { 1 }{ { \left( { \dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 } } \right) } } } \right) \\ =\left( { \dfrac { 1 }{ { \dfrac { { 8-6 } }{ { 48 } } } } } \right) +\left( { \dfrac { 1 }{ { \dfrac { { 3-2 } }{ 6 } } } } \right) \\ =\left( { \dfrac { { 48 } }{ 2 } } \right) +\left( { \dfrac { 6 }{ 1 } } \right) \\ =24+6 \\ =30 \end{array}$

Solved.

Solve
$\dfrac { 5.6\times { 10 }^{ 6 } }{ { 3\times 10 }^{ -4 }(1.76\times { 10 }^{ 11 }) }$

$r =\frac{5.6\times10^{6}}{3\times 10^{-4}(1.76\times10^{11})}$

$=\frac{5.6\times10^{6}\times10^{4}}{3\times1.76\times10^{11}}$

$=\frac{5.6\times10^{10}}{3\times1.76\times10^{11}}$

$=\frac{5.6}{3\times 1.76\times 10}$

$=\frac{5.6}{52.8}$

=0.106(approx)

1190069_1330219_ans_0ddb905305cf455dae47c9190de421c8.JPG

Find the product: $a^{2} \times 2a^{22} \times 4a^{26}$ .

$\begin{array}{l} { a^{ 2 } }\times 2{ a^{ 22 } }\times 4{ a^{ 26 } } & \left[ { { a^{ x } }.{ a^{ y } }={ a^{ x+y } } } \right] \\ = (2\times 4){ a^{ \left( { 2+22+26 } \right) } } & \\= 8{ a^{ 50 } }. & \end{array}$

Solve : $\sqrt { \dfrac { 256 a ^ { 4 } b ^ { 4 } } { 625 a ^ { 6 } b ^ { 2 } } } = ?$

$\begin{array}{l} We\, have \\ \sqrt { \dfrac { { 256{ a^{ 4 } }{ b^{ 4 } } } }{ { 625{ a^{ 6 } }{ b^{ 2 } } } } } \\ =\sqrt { \dfrac { { { { \left( { 16 } \right) }^{ 2 } }{ { \left( { { a^{ 2 } } } \right) }^{ 2 } }{ { \left( { { b^{ 2 } } } \right) }^{ 2 } } } }{ { { { \left( { 25 } \right) }^{ 2 } }{ { \left( { { a^{ 3 } } } \right) }^{ 2 } }{ { \left( b \right) }^{ 2 } } } } } \\ =\dfrac { { 16\times { a^{ 2 } }\times { b^{ 2 } } } }{ { 25\times { a^{ 3 } }\times b } } \\ =\dfrac { { 16b } }{ { 25a } } . \\ It\, is\, the\, required\, answer. \end{array}$

$\left( { 3 }^{ 4 } \right) ^{ 3 }$

Using the formula

${\left({a}^{m}\right)}^{n}={a}^{mn}$

Let

$a=(3^4)^3$

$=3^{4\times 3}$

$a=3^{12}$ .

$\therefore (3^4)^3=3^{12}$

Find the product : $\left( \dfrac { 1 } { 2 } p ^ { 3 } q ^ { 6 } \right) \left( - \dfrac { 2 } { 3 } p ^ { 4 } q \right) \left( p q ^ { 2 } \right).$

We have,

$\left(\dfrac{1}{2}p^3q^6\right)\left(-\dfrac{2}{3}p^4q\right)\left(pq^2\right)$

$\Rightarrow -\left(\dfrac{1}{3}p^8q^9\right)$

Hence, this is the answer.

If ${ (0.2) }^{ x }=2$ and ${ \log }_{ 10 }2=0.3010$ , then what is the value of $x$

$(0.2)^{x}=2$

Taking log on both sides,

$\log (0.2)^{x}=\log 2$

$x\log 0.2=0.3010 \quad \dots ( n \log_ab=\log_ab^n )$

$x(log2-log10)=0.3010 \quad \dots (\log \left(\dfrac{a}{b}\right)=\log a-\log b)$

$x(0.3010-1)=0.3010$

$x(-0.6990)=0.3010$

$x=\dfrac{-3010}{6990}=-0.43$

$x=-0.43$

$y={ log }_{ 10}x$ then $x=$

From properties of logarithm

$\log _ab=x\implies a^x=b$

Here

$a=10,b=x,x=y$

$y=\log _{10}x\implies x=10^y$

Evaluate $\log_2 128+\log _3 243$

$\log _2128+\log_3243\\\log_22^7+\log_33^5\\7\log_22+5\log_33\\7+5=12$

Evaluate:\log_{\frac{1}{100}}{\dfrac{1}{10000}}

Let

$a=\log_{\frac{1}{100}}{\dfrac{1}{10000}}$

$=\log_{{10}^{-2}}{{10}^{-4}}$

We know that,

$\log_{{a}^{n}}{{a}^{m}}=\dfrac{m}{n}\log_{a}{a}$

$a=\dfrac{-4}{-2}\log_{10}{10}$

We know that,

$\log_{a}{a}=1$

$a=2$

$\therefore \log_{\frac{1}{100}}{\dfrac{1}{10000}}=2$

${ \left( \frac { -1 }{ 2 } \right) }^{ 2 }=\frac { 1 }{ { 2 }^{ m } }$

$(-\frac{1}{2})^{2}=\frac{1}{2^{m}}$

$\frac{1}{2^{2}}=\frac{1}{2^{m}}$

$\Rightarrow 2^{m}=2^{2}$

$\therefore m=2$

1216236_1402584_ans_fa70c4e396044d37ba68f13678ab3e99.JPG

Evaluate: $\log_{49}{343}$

Let

$a=\log_{49}{343}$

$=\log_{{7}^{2}}{{7}^{3}}$

We know that,

$\log_{{a}^{n}}{{a}^{m}}=\dfrac{m}{n}\log_{a}{a}$

$a=\dfrac{3}{2}\log_{7}{7}$

Since

$\log_{a}{a}=1$

$a=\dfrac{3}{2}$

$\log_{49}{343}=\dfrac{3}{2}$

Show that $\log{\left(\dfrac{243}{343}\right)}=5\log{3}-3\log{7}$

L.H.S

$=\log{\left(\dfrac{243}{343}\right)}$

$=\log{\left(\dfrac{{3}^{5}}{{7}^{3}}\right)}$

We know that

$\log{\left(\dfrac{a}{b}\right)}=\log{a}-\log{b}$

$=\log{\left({3}^{5}\right)}-\log{\left({7}^{3}\right)}$

We know that

$\log{{a}^{m}}=m\log{a}$

$=5\log{3}-3\log{7}$

$=$ R.H.S

$\therefore \log{\left(\dfrac{243}{343}\right)}=5\log{3}-3\log{7}$

Hence, proved.

Solve:
$3^{x+1}=27\times 3^{4}$

1226847_1397929_ans_459df0250a6540c9ad70e5bcd68d5f7d.jpg

Solve: $\dfrac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}$

$\dfrac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}=\dfrac{5^2\times t^{-4}}{5^{-3}\times 5\times 2\times t^{-8}}=\dfrac{5^{2+3-1}\times t^{-4+8}}{2}=\dfrac{625}{2} \times t^{4}$

Simplify
${ 32 }^{ 1 }={ 2 }^{ x }$

$32^1=2^x$

$\Rightarrow (2)^5=2^x$

$\Rightarrow$ Comparing powers

$x=5$ .

1205630_1449175_ans_42fca8e55fcd4f54a04bd26068d95bb7.jpg

Simplify $\sqrt [3]{4}\times \sqrt [3]{16}$

$\sqrt [3]{4}\times \sqrt [3]{16}=\sqrt[3]{4\times 16}=\sqrt[3]{4\times 4^2}=\sqrt[3]{4^3}=4$

Simplify : $\dfrac{2^3 .3^4. 4}{3 .32}$

$\dfrac{2^3\times 3^4\times 4}{3\times 32}=\dfrac{2^6\times 3^4}{3\times 2^5}=2\times 3^3=2\times 27=54$

Express each of the following exponential expressions as a rational number. ${ \left( \dfrac { 2 }{ 3 } \right) }^{ \left( -1 \right) }+{ \left( \dfrac { 3 }{ 2 } \right) }^{ \left( -2 \right) }$

$\left(\dfrac{2}{3}\right)^{-1}+\left(\dfrac{3}{2}\right)^{-2}$

$\Rightarrow \dfrac{3}{2}+\left(\dfrac{2}{3}\right)^2$

$\Rightarrow \dfrac{3}{2}+\dfrac{4}{9}$

$\Rightarrow \dfrac{27+8}{18}=\dfrac{35}{18}$ .

1209712_1462226_ans_c155ecb7ed5645bab20d46a4c9452362.jpg

For $a=27, b=8, m=\dfrac{1}{3}$ verify $(ab)^{m}=a^{m}b^{m}$

$a = 27, \; b = 8, \; m = \cfrac{1}{3}$

Now,

${\left( ab \right)}^{m} = {a}^{m} {b}^{m}$

${\left( 27 \times 8 \right)}^{\tfrac{1}{3}} = {\left( 27 \right)}^{\tfrac{1}{3}} \times {\left( 8 \right)}^{\tfrac{1}{3}}$

${\left( {3}^{3} \times {2}^{3} \right)}^{\tfrac{1}{3}} = {\left( {3}^{3} \right)}^{\tfrac{1}{3}} \times {\left( {2}^{3} \right)}^{\tfrac{1}{3}}$

${\left( {\left( 2 \times 3 \right)}^{3} \right)}^{\tfrac{1}{3}} = {3}^{3 \times \tfrac{1}{3}} \times {2}^{3 \times \tfrac{1}{3}}$

${\left( 6 \right)}^{3 \times \tfrac{1}{3}} = 2 \times 3$

$6 = 6$

L.H.S.

$=$ R.H.S.

Hence verified.

Write each of the following number in the form $k \times { 10 }^{ n }$ where $1\le k\ 10$ and n is an integer.
$(v)\ 0.00729$

$0.00729$

$=7.29\times 10^{-3}$ .

1215199_1462144_ans_50d2ecaecb5344588221152546f3acdc.jpg

Solve : ${ 9 }^{ \dfrac { 5 }{ 2 } }-3\times { 8 }^{ \circ }-\left( \dfrac { 1 }{ 81 } \right) ^{ -\dfrac { 1 }{ 2 } }$

$9^{\dfrac{5}{2}}-3\times 8^o-\left(\dfrac{1}{81}\right)^{\dfrac{-1}{2}}$

$\Rightarrow (9^{1/2})^5-3\times 1-(81)^{1/2}$

$\Rightarrow 3^5-3-9$

$\Rightarrow 243-12$

$\Rightarrow 231$ .

1205686_1451627_ans_29f36c853f844d5d93f82a778e3dd1f7.jpg

$\dfrac { log\sqrt { 8 } }{ log8 }$ is equal to.

$=\dfrac{\log{\sqrt{8}}}{\log{8}}$

$=\log_{8}{\sqrt{8}}$

$=\log_{8}{{8}^{\frac{1}{2}}}$

$=\dfrac{1}{2}\log_{8}{8}$

$=\dfrac{1}{2}$

Multiply ${a^2}$ with $\left( {{a^3} + 3{a^2}b + {b^3} + 3a{b^2}} \right)$ .

By Indices Law,

$a^m \times a^n = a^{m +n}$

therefore,

$= (a^3 + 3 a^{2} b + b^3 + 3ab^2) \times a^2$

$= a^5 + 3a^4b + a^2b^3 + 3a^3b^2$

The value of $p^3$ if $13p=69$

Given relation is

$13p=69$

$p=\dfrac{69}{13}=3$

$p^3=3^3=27$

Find the value of $\log 4.5$ .

$log (4.5) = log (\frac{9}{2}) = log (9) - log (2)$

$= log (3^{2}) - log (2)$

$= 2 log (3) - log (2)$

$= 2 \times 0.477 - 0. 301$

$log (4.5) = 0.653$

Simplify:
${ \left( \dfrac { 1 }{ 2 } \right) }^{ 5 }{ -\left( \dfrac { 3 }{ 2 } \right) }^{ 3 }$

Let

$a=\displaystyle \left ( \frac{1}{2} \right )^{5}-\left ( \frac{3}{2} \right )^{3}$

$=\displaystyle\frac{1}{2^{5}}-\left ( \frac{3}{2} \right )^{3}$

$=\displaystyle\frac{1}{32}-\frac{27}{8}$

$=\displaystyle\frac{1-108}{32}$

$a=\displaystyle\frac{-107}{32}$

$\therefore \displaystyle \left ( \frac{1}{2} \right )^{5}-\left ( \frac{3}{2} \right )^{3}=\displaystyle\frac{-107}{32}$

If $a = 3$ and $b = -2$ , find the value of :
$(a+b)^{ab}$

Given,

$a=3,b=-2$

$(a+b)^{ab}$

$=(3+(-2))^{3 \times -2}$

$=1^{-6}$

$=1$

If $a = 3$ and $b = -2$ , find the value of :
$a^{b}+b^{a}$

Given,

$a=3,b=-2$

$a^b+b^a$

$=3^{-2}+(-2)^3$

$=\left ( \dfrac{1}{3} \right )^2+(-8)$

$=\dfrac{1}{9}-8$

$=-\dfrac{71}{9}$

Simplify:
${2}^{3}\times 5$

${2}^{3}\times 5=2\times 2\times 2\times 5=40$

Simplify:
$0\times{10}^{2}$

The given expression is

$0\times{10}^{2}$

$=0\times 10\times 10$

$=0$

Convert the following to logarithmic form:
$2^{6} = 64$

$2^{6} = 64$
Let us apply log, we get

$\log _{2} 64 = 6$

Convert the following to logarithmic form:
$7^{x} = 100$

$7^{x} = 100$
Let us apply log, we get

$\log _{7} 100 = x$

Convert the following to logarithmic form:
$(81)^{3/4} = 27$

$(81)^{3/4} = 27$
Let us apply log, we get

$\log _{81} 27 = 3/4$

Convert the following to logarithmic form:
$9^{0} = 1$

$9^{0} = 1$
Let us apply log, we get

$\log _{9} 1 = 0$

Convert the following to logarithmic form:
$3^{-2} = 1/9$

$3^{-2} = 1/9$
Let us apply log, we get

$\log _{3} 1/9 = -\log_{3} 9= -2$

Convert the following to logarithmic form:
$6^{1} = 6$

$6^{1} = 6$
Let us apply log, we get

$\log _{6} 6 = 1$

Convert the following to logarithmic form:
$10^{-2} = 0.01$

$10^{-2} = 0.01$
Let us apply log, we get

$\log _{10} 0.01 =\log_{10}\dfrac{1}{100}=-\log_{10}100= -2$

Convert the following to logarithmic form:
$5^{2} = 25$

$5^{2} = 25$
Let us apply log, we get

$\log _{5} 25 = 2$

Convert the following into exponential form:
$\log _{8} 4 = 2/3$

$\log _{8} 4 = 2/3$
The exponential form of the expression is

$(8)^{2/3} = 4$

Evaluate the following:
$2\log 5 + \log 8 - 1/2 \log 4$

$2\log 5 + \log 8 - 1/2 \log 4$
Let us simplify the expression,

$2 \log 5 + \log 8 - {1/2} \log 4$

$= \log 5^{2} + \log 8 - {1/2} \log 2^{2}$

$= \log 25 + \log 8 - {1/2}\times 2\log 2$

$= \log 25 + \log 8 - \log 2$

$= \log (25 \times 8)/2$

$= \log (25 \times 4)$

$= \log 100$

$= \log 10^{2}$

$= 2 \log 10$

$= 2(1)$

$=2$

Convert the following into exponential form:
$\log _{10} (0.001) = -3$

$\log _{10} (0.001) = -3$
The exponential form of the expression is

$10^{-3} = 0.001$

Evaluate the following:
$2 \log 5+ \log3 + 3 \log 2 -1/2 \log 36 - 2 \log 10$

$2 \log 5+ \log3 + 3 \log 2 -1/2 \log 36 - 2 \log 10$
Let us simplify the expression,

$= 2\log 5+\log 3+3 \log 2-1/2 \log6^2-2\log(10)$

$=2\log5+\log3 +3 \log2-1/2\times2 \log6-2 \times (1)$

$=2\log 5+\log3 +3\log2-\log(3\times2)-2$

$=2\log5+\log3+3\log2-(\log3+\log2)-2$

$=2\log5+\log3+3\log2-\log3-\log2-2$

$=2\log5+2\log2-2$

$=2\times(\log5+\log2)-2$

$=2\times\log(5\times2)-2$

$=2\log10-2$

$=2(1)-2$

$=0$

Convert the following into exponential form:
$\log _{2} 32 = 5$

$\log _{2} 32 = 5$
The exponential form of the expression is

$2^{5} = 32$

Convert the following into exponential form:
$\log _{8} 32 = 5/3$

$\log _{8} 32 = 5/3$
The exponential form of the expression is

$(8)^{5/3} = 32$

Convert the following into exponential form:
$\log _{2} 0.25 = -2$

$\log _{2} 0.25 = -2$
The exponential form of the expression is

$2^{-2} = 0.25$

Convert the following into exponential form:
$\log _{3} 81 = 4$

$\log _{3} 81 = 4$
The exponential form of the expression is

$3^{4} = 81$

Convert the following into exponential form:
$\log _{a} (1/a) = -1$

$\log _{a} (1/a) = -1$
The exponential form of the expression is

$a^{-1} = 1/a$

Convert the following into exponential form:
$\log _{3} 1/3 = -1$

$\log _{3} 1/3 = -1$
The exponential form of the expression is

$3^{-1} = 1/3$

Given $3 \left ( \log 5 - \log3 \right ) - \left ( \log 5 -2\log 6 \right ) = 2 -\log n,$ Find n .

Given :

$3 (\log 5 - \log 3) - (\log 5 - 2 \log 6) = 2 - \log n$
Let us simplify the given expression to find n,

$3\log 5 - 3\log 3 - \log 5 + 2 \log 6 = 2 - \log n$

$2 \log 5 - 3 \log 3 + 2 \log 6 = 2 (1) - \log n$

$\log 5^{2} - \log 3^{3} + \log 6^{2} = 2 \log 10 - \log n[Since, 1 = \log 10]$

$\log n = - \log 25 + \log 27 - \log 36 + \log 100$

$= (\log 100 + \log 27) - \log 5 + \log 36)$

$= \log (100 \times 27) - \log (25 \times 36)$

$= \log (100 \times 27)/(25 \times 36)$

$\log n = \log 3$

$n = 3$

Evaluate the following :
$\log 2 + 16 \log 16/15 +12 \log 25/24 + 7 \log 81/80$

we know that :

$\log mn = \log m + \log n$
and

$\log m/n = \log m - \log n$
and

$\log m^{n} = n \log m$
and

$2$ raised to power

$4 = 16$

$2$ raised to power

$3 = 8$

$3$ raised to power

$4 = 81$

$5$ raised to power

$2 = 25$

here,

$\log 2 + 16 \log 16/15 + 12 \log 25/24 + 7 \log 81/80$

$= \log 2 + 16( \log 16 - \log 15 ) + 12 ( \log 25 - \log 24) + 7 ( \log 81 - \log 80 )$

$=\log 2 + 16 \log 16 - 16 \log 15 + 12 \log 25 - 12 \log 24 + 7 \log 81 -7 \log 80$

$=\log 2 + 16 \log 16 - 16 \log 5\times3 + 12 \log 25 - 12 \log 8\times3 + 7 \log 81 -7 \log 10 \times 8$

$= \log 2 + 16 \times 4 \log 2 - 16(\log 5 + \log 3) + 12\times2 \log 5 - 12 (\log 8 +\log 3)+ 7\times 4 \log 3 - 7 (\log 10 + \log 8 )$

$=\log 2 + 64 \log 2 - 16 \log 5 - 16 \log 3 + 24 \log 5 - 12 \log 8 - 12 \log 3 +28 \log 3 - 7 \log 10 - 7 \log 8$

$= \log 2 + 64 \log 2 - 16 \log 5 - 16 \log 3 + 24 \log 5 - 12\times3 \log 2 - 12 \log 3 +28 \log 3 - 7 \log 10 - 7\times3 \log 2$

$= \log 2 + 64 \log 2 - 16 \log 5 - 16 \log 3 + 24 \log 5 - 36 \log 2 - 12 \log 3 +28 \log 3 - 7 \log 10 - 21 \log 2$

$= 65 \log 2 - 57 \log 2 + 28 \log 3 - 28 \log 3 + 24 \log 5 - 16 \log 5 - 7 \log 10$

$=8 \log 2 +8 \log 5 - 7 \log 10$

$=8 ( \log 5 + \log 2) - 7 \log 10$

$=8 \log 5 \times 2 - 7 \log 10$

$= 8 \log 10 - 7 \log 10$

$= \log 10 = 1$

Express each of the following as a single logarithm:
$1/2 \log 36 +2 \log 8 - \log 1.5$

$1/2 \log 36 +2 \log 8 - \log 1.5$
Let us simplify the expression into single logarithm,

$1/2 \log 36 +2 \log 8 - \log 1.5 = \log 36^{1/2} + \log 8^{2} - \log 1.5$

$= \log (6)^{2 \times 1/2} + \log 8^2 - \log (15/10)$

$= \log 6 + \log 64 - (\log 15 - \log 10)$

$= \log (6 \times 64) - \log 15 + \log 10$

$= \log (6\times 64\times 10) - \log 15$

$= \log [(6\times 64\times 10)/15]$

$= \log (4\times 64)$

$= \log 256$

Express the following as a single logarithm:
$2 \log _{10} 5-\log_{10} 2 + 3 \log _{10} 4 + 1$

$2 \log _{10} 5-\log_{10} 2 + 3 \log _{10} 4 + 1$
Let us simplify the expression into single logarithm,

$2 \log _{10} 5-\log_{10} 2 + 3 \log _{10} 4 + 1 = \log _{10} 5^{2} - \log _{10}2 + \log _{10} 4^{3} + \log _{10}10$

$= \log _{10} 25 - \log _{10}2 + \log _{10}64 + \log _{10}10$

$= \log _{10} (25 \times 64\times 10) - \log _{10}2$

$= \log _{10}(160000) - \log _{10}2$

$= \log _{10} (16000/2)$

$= \log _{10}8000$

Evaluate the following :
$2 \log _{10} 5 + \log_{10}8 -1/2 \log_{10} 4$

$2 \log_{10} 5 +\log_{10} 8 - (1/2) \log_{10} 4$

$=\log_{10} 5^2 + \log_{10} 8 - \log_{10} 4^{1/2}$

$=\log_{10} 25 + \log_{10} 8 - \log_{10} 2$

$=\log_{10} 25 + \log_{10} (8/2)$

$=\log_{10} 25 + \log_{10} 4$

$=\log_{10} (25\times4)$

$=\log_{10} 100$

$=\log_{10} 10^2$

$=2\log_{10} 10$

$=2$

Solve for $x:$
$\log x + \log 5=2 \log 3$

$\log z + \log 5 = 2 \log 3$
Let us solve for

$x$ ,

$\log x = 2 \log 3 - \log 5$

$= \log 3^{2}- \log 5$

$= \log 9 - \log 5$

$= \log (9/5)$

$\therefore x = 9/5$

Express each of the following as a single logarithm:
$1/2 \log 25-2 \log 3+1$

$1/2 \log 25-2 \log 3+1$
Let us simplify the expression into single logarithm,

$1/2 \log 25-2 \log 3+1 = \log 25^{1/2} - \log 3^{2} + \log 10$

$= \log (5)^{2\times1/2} - \log 9 + \log 10$

$= \log 5 - \log 9 + \log 10$

$= \log (5 \times 10) - \log 9$

$= \log ((5 \times 10)/9)$

$= \log 50/9$

Express each of the following as a single logarithm:
$2 \log 3 - 1/2 \log 16 + \log 12$

$2 \log 3 - 1/2 \log 16 + \log 12$
Let simplify the expressions into single logarithm,

$2 \log 3 - 1/2 \log 16 + \log 12 = 2 \log 3 - ^{1/2} \log 4^{2} + \log 12$

$= 2 \log 3 - \log 4 + \log 12$

$= \log 3^{2} - \log 4 + \log 12$

$= \log (9\times 12)/4$

$= \log (9\times 3)$

$= \log 27$

Prove the following :
$\log 4\div \log_{10} 2= \log_{3} 9$

$\log 4\div \log_{10} 2= \log_{3} 9$
Let us consider LHS,

$\log 4\div \log_{10} 2$

$\log _{10}4 \div \log _{10}2 = \log _{10} 2^{2} \div \log _{10}2$

$= 2 \log _{10} 2 \div \log _{10}2$

$= 2 \log _{10} 2/ \log _{10}2$

$= 2 (1)$

$= 2$

Prove the following :
$\log_{10} 25 + \log_{10} 4 = \log_{5} 25$

$\log_{10} 25 + \log_{10} 4 = \log_{5} 25$
Let us consider LHS,

$\log_{10} 25 + \log_{10} 4= \log _{10} (25 \times 4)$

$= \log _{10} 100$

$= \log _{10} 10^{2}$

$= 2 \log _{10} 10$

$= 2(1)$

$= 2$

And RHS,

$\log_{5}25$

$=\log_{5}5^2$

$=2\log_{5}5$

$=2(1)$

$=2$

Solve for $x$ :
$x = \log 125/ \log 25$

$x = \log 125/\log 25$

$x = \log 5^{3} / \log 5^{2}$

$= 3 \log 5/2 \log 5$

$= 3/2$ [Since,

$\log 5/\log 5 = 1$ ]

$\therefore x = 3/2$

Solve the following equations :
$\log \left ( 3x +2 \right )+ \log\left ( 3x - 2 \right ) = 5 \log 2$

$\log \left ( 3x +2 \right )+ \log\left ( 3x - 2 \right ) = 5 \log 2$
Let us simplify the expression,

$\log (3x + 2) + \log (3x - 2) = \log 2^{5}$

$\log [(3x + 2) \times (3x -2)] = \log 32$

$\log (9x^{2} - 4) = \log 32$

$(9x^{2} - 4) = 32$

$9x^{2} = 32 + 4$

$9x^{2} = 36$

$x^{2} = 36/9$

$x^{2} = 4$

$x = \sqrt{4}$

$x= 2$

Prove the following :
$27 ^{\log 2 } = 8^{\log 3}$

$27 ^{\log 2 } = 8^{\log 3}$
Let us take

$\log$ on both sides,
If

$27 ^{\log 2 } = 8^{\log 3}$

$\log 2 . \log 27 = \log 3 . \log 8$

$\log 2 . \log 3^{3} = \log 3 . \log 2^{3}$

$\log 2 . 3 \log 3 = \log 3 . 3 \log 2$

$3 \log 2 . \log 3 = 3\log 2 . \log 3$
Which is true.
Hence proved.

Solve for x:
$\log_{3}x- \log_{3} 2=1$

$\log_{3}x- \log_{3} 2=1$
Let us solve for x,

$\log_{3} x = 1 + \log_{3} 2$

$= \log_{3} 3 + 2$ [Since, 1 can be written as

$\log_{3} 3 = 1$ ]

$= \log_{3} (3 \times 2)$

$= \log_{3} 6$

$\therefore x = 6$

If $\log x /\log 5 = \log y^{2} / \log 9/ \log \left ( 1/3 \right ),$ find $x$ and $y$ .

Given :

$\log x / \log 5 = \log y^{2}/ \log 2 = \log 9/v(1/3)$
Let us consider,

$\log x/\log 5 = \log 9 /\log (1/3)$

$\log x = (\log 9 \times \log 5)/\log (1/3)$

$= (\log 3^{2} \times \log 5)/(\log 1 = \log 3)$

$= (2 \log 3\times \log 5)/(-\log 3)[\log 1 = 0]$

$= -2 \times \log 5$

$= \log 5^{-2}$

$x = 5^{-2}$

$= 1/5^{2}$

$= 1/25$
Now,

$\log y^{2}/\log 2 = \log 9/ \log (1/3)$

$\log y^{2}= (\log 9 \times \log 2)/\log (1/3)$

$= (\log 3^{2} \times \log 2)/(\log 1 - \log 3)$

$= (2\log 3 \times \log 2)/(-\log 3)[\log 1 = 0]$

$= -2 \times \log 2$

$= \log 2^{-2}$

$y^{2} = 2^{-2}$

$= 1/2^{2}$

$= 1/4$

$y= \sqrt{(1/4)}$

$= 1/2$

Given $2 \log _{10} x+1 = \log _{10} 250,$ find $x$

Given :

$2 \log_{10} x + 1 =\log_{10} 250$
Let us simplify the above expression,

$\log_{10} x^{2} + \log_{10} 10 = \log_{10} 250$ [Since, 1 can be written as

$\log_{10} 10$ ]

$\log_{10} (x^{2} \times 10) = \log_{10} 250$

$x^{2} = 250/10$

$x^{2} = 25$

$x = \sqrt{25}$

$\therefore x = 5$

Prove the following :
$3 ^{\log 4 } = 4^{\log 3}$

$3 ^{\log 4 } = 4^{\log 3}$
Let us take log on both sides,
If

$3 ^{\log 4 } = 4^{\log 3}$

$\log 4 . \log 3 = \log 3 . \log 4$

$\log 2^{2} . \log 3 = \log 3 . \log 2^{2}$

$2 \log 2 . \log 3 = \log 3 . 2 \log 2$
Which is true.
Hence proved

Solve for $x:$
$\left ( \log 8/\log 2 \right ) \times \left ( \log 3/\log \sqrt{3} \right ) = 2 \log x$

$\left ( \log 8/\log 2 \right ) \times \left ( \log 3/\log \sqrt{3} \right ) = 2 \log x$

$(\log 2^{3}/ \log 2) \times (\log 3 / \log 3^{1/2}) = 2 \log x$

$(3\log 2 / \log 2) \times (\log 3/1/2\log 3) = 2 \log x$

$3 \times 1/(1/2) = 2\log x$

$6 = 2\log x$

$\log x = 6/2$

$\log x = 3$

$x= 10^3$

$\therefore x = 1000$

Solve the following equations :
$\log_{10}\left ( x +2 \right ) + \log_{10}\left ( x-2 \right ) = \log_{10}3+3 \log_{10}4$

$\log_{10}\left ( x +2 \right ) + \log_{10}\left ( x-2 \right ) = \log_{10}3+3 \log_{10}4$
Let us simplify the expression,

$\log_{10}\left ( x +2 \right ) + \log_{10}\left ( x-2 \right ) = \log_{10}3+3 \log_{10}4$

$\log_{10} [(x + 2) \times (x - 2)] = \log_{10} 3 + \log_{10} (3 \times 4^{3})$

$[(x +2) \times (x - 2)] = (3 \times 4^{3})$

$(x^{2} - 4) = (3 \times 4 \times 4 \times 4)$

$(x^{2} - 4) = 192$

$x^{2} = 192 + 4$

$= 196$

$x = \sqrt{196}$

$= 14$

Given $2 \log _{10} x+1 = \log _{10} 250,$ find
$\log _{10} 2 x$

Given :

$\log_{10}2x$

First we need to find the value of

$x$ :

$2 \log_{10} x + 1 =\log_{10} 250$
Let us simplify the above expression,

$\log_{10} x^{2} + \log_{10} 10 = \log_{10} 250$ [Since, 1 can be written as

$\log_{10} 10$ ]

$\log_{10} (x^{2} \times 10) = \log_{10} 250$

$x^{2} = 250/10$

$x^{2} = 25$

$x = \sqrt{25}$

$\therefore x = 5$

Now, we know that,

$x = 5$ So,

$\log_{10} 2x = \log_{10} 2 \times 5$

$= \log_{10} 10$

$=1$

Show that :
$1 / \log_{2} 42 + 1 / \log_{3} 42+ 1 / \log_{7} 42 = 1$

$1 / \log_{2} 42 + 1 / \log_{3} 42+ 1 / \log_{7} 42 = 1$
Let us consider LHS:

$1 / \log_{2} 42 + 1 / \log_{3} 42+ 1 / \log_{7} 42$
By using the formula,

$\log _{n} m = \log {m} / \log {n}$

$1/\log _{2} 42 + 1 / \log _{3}42 + 1/\log _{7}42 = 1/(\log 42/ \log {2}) + 1/(\log 42 / \log {3}) + 1/(\log 42/\log {7})$

$= \log {2} + \log {3} +\log {7} / \log 42$

$=\log(2\times3\times7)/\log 42$

$= \log 42 / \log 42$

$= 1$

$=$ RHS

Find out the value of $log(8621).$

The given number,

$x = 8621 = 8.621 \times 10^3$

$\therefore$ The characteristic of

$x = 3$

From log table the mantisa of

$x = 0.9356$

Therefore

$log x = log 8621 = 3.9356$

Solve for x:
$\log_{2}x+ \log_{8}x+ \log_{32} x = 23/15$

$\log_{2}x+ \log_{8}x+ \log_{32} x = 23/15$
let us simplify the expression,

$1/\log_{x} 2 + 1/\log_{x} 8 + 1/\log_{x} 32 = 233/15$

$1/\log_{x} 2^{1} + 1/\log_{x} 2^{3} + 1/\log_{x} 2^{5} = 23 / 15$

$1/\log_{x} 2 + 1/3\log_{x} 2 + 1/5\log_{x} 2 = 23/15$

$1/\log_{x} x [(15 + 5 + 3)/ 15] = 23/15$

$\log_{2} x [23/15] = 23/15$

$\log_{2} x = (23/15) \times (15/23)$

$\log_{2} x = 1$

$\log_{2} x = \log_{2} 2$ [Since,

$1 = \log_{a}a]$

$x=2$

Use logarithm table to find the logarithm of the following numbers:
$25795$

We have to find

$\log 25795$

Firstly, we will write

$25795$ in standard form

$2.5795 \times 10^4$

So, the characteristic of the logarithm of 25795 is 4.

To find the mantissa of the logarithm of 25795, we take the first four digits.
The number formed by the first four digits is 2579.

Now, we look in the row starting with 25. In this row, look at the number in the column headed by 7. The number is 4099.

Now, move to the column of mean differences and look under the column headed by 9 in the row corresponding to 25. We see that the number there is 15.

Add this number to 4099. We get the number 4114. This is the required
mantissa.

Hence,

$\log(25795)=4.4114$

Ans: 4.4114

Using logarithm, find the value of $6.45\times 981.4$

Let

$x=6.45\times 981.4$ .

Taking log on both sides

$\log x=\log \left ( 6.45\times 981.4 \right )$

$=\log 6.45+\log 981.4$

$=0.8096+2.9919$ (using log table)

$=3.8015$

$\therefore x=antilog (3.8015)=6331.0$ (using antilog table)

Ans: 6331

Given that $\displaystyle \log_a A = x$ is similar to $\displaystyle a^x = A$ .
If true then write 1 and if false then write 0

Given,

$\log_aA=x$

$\implies \log$ of

$x$ to the base

$a$ is

$A$

$\implies A=a^x$ .

$\log 3.6$ is $0.55a2$ . where a is 3rd digit of the given value,then $a=?$

$x= \log 3.6$

$\therefore x = \log 36 - \log 10$

$\therefore x = 2 \log (2 \cdot 3) - 1$ ...

$\log a^b = b \log a$

$\therefore x = 2\log 2 + 2 \log 3 - 1$ ..

$\log (a \cdot b) = \log a + \log b$

$\therefore x = 2 \log 2 + 2 \log 3 - 1$ ...(

$\log a^b = b \log a$ )

$\therefore x= 2\times (0.3010 + 0.4771) = 0.5562$

If $\displaystyle \log_3 m = x$ and $\displaystyle \log_3 n = y$ , then $\displaystyle 3^{1 - 2y + 3x}$ can be expressed in terms of $m$ and $n$ as $\displaystyle \frac {3m^3}{n^2}$ .
If true then write 1 and if false then write 0.

Given,

$\displaystyle \log_3 m = x$ and

$\log_3n = y$

$\therefore m=3^x$ and

$n=3^y$ ...(1)
Thus,

$\displaystyle 3^{1 - 2y + 3x}$

$= \dfrac{3\times 3^{3x}}{3^{2y}}$

$\therefore \dfrac{3\times 3^{3x}}{3^{2y}}= \dfrac{3m^3}{n^2}$ (From 1)

$\displaystyle 10^a$ in terms of $x$ is $\displaystyle \sqrt x$ .

If true then write 1 and if false then write 0.

Given,

$\log _{ 10 }{ x } = 2a$

$x = { 10 }^{ 2a }$

Taking square root on both side,

$\sqrt { x } = { 10 }^{ a }$

${ 10 }^{ a }$ in terms of

$x$ is

$\sqrt x$ .

$\displaystyle \log _{2}32= 5$

$\textbf{Step 1:Definition of logarithm:}\ \boldsymbol{\log_{a} x = b \quad \Leftrightarrow\ a^{b} = x}$ , (

$\textbf{Where a , x > 0 & a }\neq 1$ )

$\textbf{Given,}$

$\log_{2} 32 = 5 \quad \Rightarrow\,\, 32 = 2^{5}$ (

$\because \boldsymbol{\log_{a} x = b \quad \Leftrightarrow\ a^{b} = x}$ )

$2^{5} = 32$ (

$\textbf{Exponent form}$ )

Evaluate: ${ \left( 6.32 \right) }^{ 2 }\times \sqrt [ 4 ]{ 83.94 }$
[Hint: Use logarithm tables]

Let

$y={ (6.32) }^{ 2 }\times \sqrt [ 4 ]{ 83.94 }$

Taking

$\log$ on both the sides, we get

$\ln { y } =\ln { \left( { (6.32) }^{ 2 }\times \sqrt [ 4 ]{ 83.94 } \right) }$

$\Rightarrow \ln { y } =\ln { { (6.32) }^{ 2 } } +\ln { \sqrt [ 4 ]{ 83.94 } }$

$\Rightarrow \ln { y } =2\ln { 6.32 } +\dfrac { 1 }{ 4 } \ln { 83.94 }$

$\Rightarrow \ln { y } =3.687+1.107$

$\Rightarrow \ln { y } =4.794$

$\Rightarrow y=$ anti

$\ln { 4.794 }$

$\Rightarrow y=120.88$

which can be approximated to

$121$ .

Evaluate: $\dfrac { { \left( 17.42 \right) }^{ \frac { 2 }{ 3 } }\times 18.42 }{ \sqrt { 126.37 } }$
If the answer is not an whole number then write the whole number just smaller than the answer.
[Hint: Use logarithm tables]

Let

$y=\dfrac { { (17.42) }^{ \frac { 2 }{ 3 } }\times 18.42 }{ \sqrt { 126.37 } }$

Taking

$\log$ on both sides, we get

$\ln { y } =\ln { \left( \dfrac { { (17.42) }^{ \frac { 2 }{ 3 } }\times 18.42 }{ \sqrt { 126.37 } } \right) }$

$\Rightarrow \ln { y } =\ln { { (17.42) }^{ \frac { 2 }{ 3 } } } +\ln { 18.42 } -\ln { \sqrt { 126.37 } }$

$\Rightarrow \ln { y } =\dfrac { 2 }{ 3 } \ln { (17.42) } +\ln { 18.42 } -\dfrac { 1 }{ 2 } \ln { (126.37) }$

$\Rightarrow \ln { y } =1.905+2.913-2.419$

$\Rightarrow \ln { y } =2.398$

$\Rightarrow y=$ anti

$\ln { 2.398 }$

$\Rightarrow y=11.004$

So, answer is

$11$ .

Find the value of ' $x$ ' such that
$25\times {5}^{x}={5}^{8}$

$25\times {5}^{x}={5}^{8}$

$\implies {5}^{2}\times {5}^{x}={5}^{8}$

$[\because 25=5\times 5={5}^{2}]$

$\implies {5}^{2+x}={5}^{8}$

$[\because {a}^{m}\times {a}^{n}={a}^{m+n}]$

$\implies 2+x=8$

$[\because a^m=a^n \implies m=n]$

$\therefore$

$x=6$

Find the value of $\log_{81}3$

Formula used

$\displaystyle \log_{a^n}b^m=\frac mn \log_a b$

Now,

$\displaystyle \log_{81}3=\log_{3^4}3^1$

$=\cfrac 14 \log_3 3$

$=\cfrac 14$

$(\because \log_aa=1)$

Simplify:
${ 10 }^{ -1 }\times { 10 }^{ 2 }\times {10}^{-3} \times { 10 }^{ 4 }\times { 10 }^{ -5 }\times { 10 }^{ 6 }$

Given

${ 10 }^{ -1 }\times { 10 }^{ 2 }\times { 10 }^{ -3 }\times { 10 }^{ 4 }\times { 10 }^{ -5 }\times { 10 }^{ 6 }$

We know that

${x}^{m}\times{{x}^{n}}={x}^{(m+n)}$ so simplifying using this as below

${ 10 }^{ \left( 2+4+6 \right) }\times { 10 }^{ \left( -1-3-5 \right) }$

${ 10 }^{ 12 }\times { 10 }^{ -9 }$

${ 10 }^{ \left( 12-9 \right) }$

${ 10 }^{ 3 }$

Simplify $(2^{3})^{-2} \times (3^{2})^{2}$

$(2^{3})^{-2} \times (3^{2})^{2} = 2^{(3\times -2)} \times 3^{(2\times 2)}$

$= 2^{-6}\times 3^{4} = \dfrac {1}{2^{6}} \times 3^{4} = \dfrac {3^{4}}{2^{6}} = \dfrac {81}{84}$

Obtain the equivalent logarithmic form of the following.
(i) ${ 2 }^{ 4 }=16$
(ii) ${ 3 }^{ 5 }=243$
(iii) ${ 10 }^{ -1 }=0.1$
(iv) ${ 8 }^{ -\frac { 2 }{ 3 } }=\dfrac { 1 }{ 4 }$
(v) ${ 25 }^{ \frac { 1 }{ 2 } }=5$
(vi) ${ 12 }^{ -2 }=\dfrac { 1 }{ 144 }$

(i)

$\:\:\:\log_{2}{16}=4$

(ii)

$\:\:\:\log_{3}{243}=5$

(iii)

$\:\:\:\log_{10}{0.1}=-1$

(iv)

$\:\:\:\log_{8}\dfrac{1}{4}=\dfrac{-2}{3}$

(v)

$\:\:\:\log_{25}{5}=\dfrac{1}{2}$

(vi)

$\:\:\:\log_{12}\left ( \dfrac{1}{144} \right )=-2$

Simplify $\dfrac {(2^{2})^{3}}{(3^{2})^{2}}$

$\dfrac {(2^{2})^{3}}{(3^{2})^{2}} = \dfrac {2^{2\times 3}}{3^{2\times 2}} = \dfrac {2^{6}}{3^{4}} = \dfrac {64}{81}$ .

If $2^{p} = 32$ , find the value of $p \ .$

Given:

$2^{p} = 32$

$2^{p} =2\times2\times2\times2\times2= 2^{5}$

$\Rightarrow 2^{p} = 2^{5}$

Hence, the value of

$p$ is

$5.$

Find the value of the following:
$\left (\dfrac {3}{8}\right )^{5} \times \left (\dfrac {3}{8}\right )^{4}\div \left (\dfrac {3}{8}\right )^{9}$

$\left (\dfrac {3}{8}\right )^{5} \times \left (\dfrac {3}{8}\right )^{4}\div \left (\dfrac {3}{8}\right )^{9} = \dfrac {\left (\dfrac {3}{8}\right )^{5 + 4}}{\left (\dfrac {3}{8}\right )^{9}} = \dfrac {\left (\dfrac {3}{8}\right )^{9}}{\left (\dfrac {3}{8}\right )^{9}} = 1$
(or)

$\left (\dfrac {3}{8}\right )^{9 - 9} = \left (\dfrac {3}{8}\right )^{0} = 1$

Find the value of the following:
$3^{4} \times 3^{-3}$

$3^{4} \times 3^{-3} = 3^{4 + (-3)} = 3^{4 - 3} = 3^{1} = 3$

Change the following from exponential form to logarithmic form.
(i) ${ 3 }^{ 4 }=81$
(ii) ${ 6 }^{ -4 }=\dfrac { 1 }{ 1296 }$
(iii) ${ \left( \dfrac { 1 }{ 81 } \right) }^{ \frac { 3 }{ 4 } }=\dfrac { 1 }{ 27 }$
(iv) ${ \left( 216 \right) }^{ \frac { 1 }{ 3 } }=6$
(v) ${ \left( 13 \right) }^{ -1 }=\frac { 1 }{ 13 }$

(i)

${ 3 }^{ 4 }=81\Longrightarrow \log _{ 3 }{ 81 } =4$
(ii)

${ 6 }^{ -4 }=\dfrac { 1 }{ 1296 } \Longrightarrow \log _{ 6 }{ \left( \dfrac { 1 }{ 1296 } \right) } =-4$
(iii)

${ \left( \dfrac { 1 }{ 81 } \right) }^{ \frac { 3 }{ 4 } }=\dfrac { 1 }{ 27 } \Longrightarrow \log _{ \frac { 1 }{ 81 } }{ \left( \dfrac { 1 }{ 27 } \right) } =\dfrac { 3 }{ 4 }$
(iv)

${ \left( 216 \right) }^{ \frac { 1 }{ 3 } }=6\Longrightarrow \log _{ 216 }{ 6 } =\dfrac { 1 }{ 3 }$
(v)

${ \left( 13 \right) }^{ -1 }=\dfrac { 1 }{ 13 } \Longrightarrow \log _{ 13 }{ \left( \dfrac { 1 }{ 13 } \right) } =-1$

Simplify: $2^{5}\times 2^{3}$

$2^{5} \times 2^{3} = 2^{5 + 3} = 2^{8}$

Find (i) $\log { 86.76 }$
(ii) $\log { 730.391 }$
(iii) $\log { 0.00421526 }$

(i)

$86.76=8.676\times { 10 }^{ 1 }$ (scientific notation)

$\therefore$ The characteristic is

$1$ . To find mantissa consider the number

$8.676$ .
From the table,

$\log { 8.67 }$ is

$0.9380$
Mean difference of

$6$ is

$0.0003$

$\log { 8.676 } = 0.9380+0.0003=0.9383$

$\therefore \log { 86.76 } =1.9383$
(ii)

$730.391=7.30391\times { 10 }^{ 2 }$ (scientific notation)

$\therefore$ The characteristic is

$2$ . To find mantissa consider the number

$7.30391$ and approximate it as

$7.304$ (since

$9$ in the fourth decimal place is greater than

$5$ )
From the table,

$\log { 7.30 }$ is

$0.8633$
Mean difference of

$4$ is

$0.0002$

$\log { 7.304 } =0.8633+0.0002=0.8635$

$\therefore \log { 730.391 } =2.8635$
(iii)

$0.00421526=4.21526\times { 10 }^{ -3 }$ (scientific notation)

$\therefore$ The characteristic is

$-3$ . To find mantissa consider the number

$4.21526$ and approximate it as

$4.215$ (since

$2$ in the fourth decimal place is less than

$5$ ).
From the table,

$\log { 4.21 }$ is

$0.6243$
Mean difference of

$5$ is

$0.0005$

$\log { 4.215 } =0.6243+0.0005=0.6248$

$\therefore \log { 0.00421526 } =-3+0.6248=\overline { 3 } .6248$

If $Kx = |\ln (x)|$ has $3$ solutions find out the limiting values of $K$ for which this is possible.

We have to draw the graphs of both the functions.

L is

$y=k'x$ ; where

$k'$ is the limiting value of

$k$ . Below

$k'$ ; the equations will have 3 solutions for all

$x$ .

Let

$y=k'x$ touch

$y=\left | \ln(x) \right |$ at

$x=x_{o}$

$\therefore \:$ Slope of

$\left | \ln(x) \right |$ at

$x=x_{o}$ will be

$\dfrac{1}{x_{o}}$

and, Slope of

$(k'x)$ at

$x=x_{o}=k'$

Also,

$k'x_{o}=\ln(x_{o})\:\Rightarrow \: k'=\dfrac{\ln(x_{o})}{x_{o}}$

and,

$k'=\dfrac{1}{x_{o}}$

$\Rightarrow \: \dfrac{\ln(x_{o})}{x_{o}}=\dfrac{1}{x_{o}}\:\:\Rightarrow \:\: \dfrac{1}{x_{o}}\left ( \ln(x_{o})-1 \right )=0$

$\Rightarrow \: x_{o}=e$

Hence, the limiting values of

$k$ for which equation will have 3 solutions will be,

$k\:\epsilon \: (0,\dfrac{1}{e})$

Given $\log_{10} 2 = {^{.}30103},$ find $\log_{25}200.$

$\log_{25}{200}=\dfrac{\log200}{\log25}=\dfrac{\log2+\log100}{\log{\dfrac{100}{4}}}=\dfrac{\log2+2}{2-2\log{2}}=\dfrac{2.30103}{1.39794}=1.646$

Write the characteristic of each of the following
(i) $\log { 4576 }$
(ii) $\log { 24.56 }$
(iii) $\log { 0.00257 }$
(iv) $\log { 0.0756 }$
(v) $\log { 0.2798 }$
(vi) $\log { 6.453 }$

(i)

$\log{(4576)}=\log{(10^{3}\times4.576)}$

$=3+\log_{10}{4.576}\:\:\Rightarrow$ Characteristic

$=3$

(ii)

$\log{(24.56)}=\log{(10\times2.456)}=1+\log(2.456)\:\:\Rightarrow$ Characteristic

$=1$

(iii)

$\log{(2.57\times10^{-3})}=-3+\log{(2.57)}\:\:\Rightarrow$ Characteristic

$=-3$

(iv)

$\log{(0.0756)}=\log{(7.56\times10^{-2})}$

$=-2+\log{(7.56)}\:\:\Rightarrow$ Characteristic

$=-2$

(v)

$\log{(2.798\times 10^{-1})}=-1+\log{(2.798)}\:\:\Rightarrow$ Characteristic

$=-1$

(vi)

$\log{(6.453)}=0+\log{(6.453)}\:\:\Rightarrow$ Characteristic

$=0$

Write the characteristic of the following.
(i) $\log { 27.91 }$
(ii) $\log { 0.02871 }$
(iii) $\log { 0.000987 }$
(iv) $\log { 2475 }$

(i)

$\log_{10}{(10\times2.791)}=\log_{10}{10}+\log_{10}{(2.791)}$

$=1+\log_{10}{(2.791)}\Rightarrow \:$ Characteristic

$=1$

(ii)

$\log_{10}{(2.871\times 10^{-2})}=-2+\log_{10}{(2.871)}\Rightarrow \:$ Characteristic

$=-2$

(iii)

$\log_{10}{(9.87\times 10^{-4})}=-4+\log_{10}{(9.87)}\Rightarrow \:$ Characteristic

$=-4$

(iv)

$\log{(2475)}=\log{(2.475\times 10^{3})}$

$=\log_{10}{10^{3}}+\log_{10}{2.475}=3+\log_{10}{2.475}\Rightarrow \:$ Characteristic

$=+3$

Given that $\log { 4586 } =3.6615$ , find
(i) $\log { 45.86 }$
(ii) $\log { 45860 }$
(iii) $\log { 0.4586 }$
(iv) $\log { 0.004586 }$
(v) $\log { 0.04586 }$
(vi) $\log { 4.586 }$

(i)

$\log{(4586\times10^{-2})}=-2+\log{(4586)}$

$=-2+3.6615=1.6615$

(ii)

$\log{(4586\times 10)}=1+\log{(4586)}$

$=1+3.6615=4.6615$

(iii)

$\log{(4586\times 10^{-4})}=-4+\log{(4586)}$

$=-4+3.6615=0.3385$

(iv)

$\log{(4586\times10^{-6})}=-6+\log{(4586)}$

$=-6+3.6615=-2.3385$

(v)

$\log{(4586\times10^{-5})}=-5+\log{(4586)}=-1.3385$

(vi)

$\log{(4586\times10^{-3})}=-3+\log{(4580)}=0.6615$

Determine the value of x for which the 6th term in $\left ( 2^{log_2(\sqrt{g^{x-y}+7})}+\frac{1}{2^{\frac{1}{5}log_2(3^{x-1}+1)}} \right )$ is 84.

Use formula

$\Rightarrow { a^{ Log\begin{array} { *{ 20 }{ c } }c \\ a \end{array} } }={ c^{ Log\begin{array} { *{ 20 }{ c } }a \\ a \end{array} } }$

$\begin{matrix} \Rightarrow Lo{ g_{ 2 } }\sqrt { { 9^{ x-1 } }+7 } ={ \left( { \sqrt { { 9^{ x-1 } }+7 } } \right) ^{ Log\begin{array} { *{ 20 }{ c } }a \\ a \end{array} } } \\ \Rightarrow { \left( { \sqrt { { 9^{ x-1 } }+7 } } \right) ^{ 1 } }--------(i) \\ \end{matrix}$

Use formula

$\Rightarrow \log {a_a} = 1$

Now,

${2^{1/5Lo{g_2}\left( {{3^{x - 1}} + 1} \right)}} = {2^{Lo{g_2}{{\left( {{3^{x - 1}} + 1} \right)}^{1/5}}}}$

Use formula

${ a^{ Log{ c_{ a } }=c } }$

$\Rightarrow {\left( {{3^{x - 1}} + 1} \right)^{1/5}} - - - - - - - - (ii)$

$From(i)\& (ii) \Rightarrow \left( {\sqrt {{9^{x - 1}} + 7} + \frac{1}{{{{\left( {{3^{x - 1}} + 1} \right)}^{1/5}}}}} \right)$

${T_{r + 1}}{ = ^7}{c_r}{\left( {{9^{x - 1}} + 7} \right)^{7 - r/2}} \times \frac{1}{{{{\left( {{3^{x - 1}} + 1} \right)}^{r/5}}}}$

$\begin{matrix} \because { T_{ r } }=84 \\ \Rightarrow { T_{ 5+1 } }=84 \\ \Rightarrow r=5 \\ { \Rightarrow ^{ 7 } }{ c_{ 5 } }{ \left( { { 9^{ x-1 } }+7 } \right) ^{ 1 } }\times \frac { 1 }{ { { { \left( { { 3^{ x-1 } }+1 } \right) }^{ 1 } } } } =84 \\ \end{matrix}$

Using the formula

$^n{c_r} = \frac{{n!}}{{n - 1!r!}}$

$\begin{array}{l} \Rightarrow \left( { \frac { { 7\cdot 6\cdot 5! } }{ { 2!5! } } { { \left( { { 9^{ x-1 } }+7 } \right) }^{ 1 } }\times \frac { 1 }{ { { { \left( { { 3^{ x-1 } }+1 } \right) }^{ 1 } } } } } \right) =84 \\ \Rightarrow 21\left( { { 9^{ x-1 } }+7 } \right) =84\left( { { 3^{ x-1 } }+1 } \right) \\ \Rightarrow { 9^{ x-1 } }+7=4\left( { { 3^{ x-1 } }+1 } \right) \end{array}$

$\begin{array}{l} Put\, \, \, { 3^{ x } }=t,{ 9^{ x } }={ t^{ 2 } } \\ \Rightarrow \frac { { { t^{ 2 } } } }{ 5 } +7=\frac { { 4t+4 } }{ 3 } \\ \Rightarrow { t^{ 2 } }-12t+27=0 \\ \Rightarrow \left( { t-9 } \right) \left( { t-3 } \right) =0 \\ t=9\, \, \, \, or\, \, \, t=3 \\ \Rightarrow { 3^{ x } }={ 3^{ 2 } }\, \, \, \, \, \, or\, \, \, \, { 3^{ x } }={ 3^{ 1 } } \\ x=1\, \, \, or\, \, 2 \end{array}$

Solve the following equation:
$\displaystyle\, \frac{3}{2^{\log x}} = \frac{1}{64}$

$\frac { 3 }{ 2^{logx} } =\frac { 1 }{ 64 } \\ \Rightarrow { 2 }^{ logx }=(3).64\\ \Rightarrow logx={ log }_{ 2 }3+6\\ \Rightarrow x=({ 10 }^{ log_{ 2 }3 }){ 10 }^{ 6 }\\ \Rightarrow x={ 10 }^{ 6 }({ 10 }^{ log_{ 2 }3 }).$

Solve the following equations.
$\displaystyle (6.25)^{2 \, - \, x} = \, \frac{1}{2^{x \, + \, 3}}.$

${ 2 }^{ 2-x }.{ (3.25) }^{ 2-x }=\frac { 1 }{ { 2 }^{ x+3 } }$

$\Rightarrow { 2 }^{ 5 }.{ (3.25) }^{ 2-x }=1$

$\\ \Rightarrow 2-x=-5\log _{ 75 }{ 2 } \\ therefore\\ x=2+5\log _{ 75 }{ 2 }$

Solve the following equations.
$\displaystyle 6.9^{0.5x \, - \, 2} \, + \, 2.3^{x \, - \, 6} \, = \, 56.$

$6.{ 9 }^{ 0.5x-2 }+2.{ 3 }^{ x-6 }=(2.3).{ 3 }^{ 2(0.5x-2) }+2.{ 3 }^{ x-6 }$

$2.{ 3 }^{ x-5 }+2.{ 3 }^{ x-6 }=56\\ 8.{ 3 }^{ x-5 }=56.3\\$

therefore

$x=6+\log _{ 3 }{ 7 }$

Find the value of $x$ if ${2^4} \times {2^5} = {\left( {{2^3}} \right)^x}.$

${2^4} \times {2^5} = {\left( {{2^3}} \right)^x}$

$\Rightarrow {2^{4 + 5}} = {\left( {{2^3}} \right)^x}$

$\Rightarrow {2^9} = {2^{3x}}$
Comparing powers,

$\Rightarrow 3x = 9$

$\Rightarrow x = 3$

Represent the following mixed infinite decimal periodic fractions as common fractions:
$\displaystyle\, log_{1/2} (log_3\, cos(\pi /6) - log_3\, sin (\pi /6))$

We know that

$log(a)-log(b)=log(a/b)$

$\Rightarrow \log _{ 3 }{ \cos { (\dfrac { \pi }{ 6 } ) } } -\log _{ 3 }{ \sin { (\dfrac { \pi }{ 6 } ) } } =\log _{ 3 }{ \cot { \dfrac { \pi }{ 6 } } } =\log _{ 3 }{ \sqrt { 3 } } =\dfrac { 1 }{ 2 }$

$\therefore$

$\log _{ \frac { 1 }{ 2 } }{ \dfrac { 1 }{ 2 } } =1$

simplify: ${\left\{ {5\left( {{{16}^{\dfrac{1}{4}}} + {{27}^{\dfrac{1}{3}}}} \right)} \right\}^{\dfrac{1}{4}}}$

${5((2^4)^{\cfrac{1}{4}}+(3^3)^{\cfrac{1}{3}})}^{\cfrac{1}{4}}\\ \Rightarrow{5(2+3)}^{\cfrac{1}{4}}\\ \Rightarrow(5^2)^{\cfrac{1}{4}}\\ \Rightarrow\sqrt5$

Solve
$\log {10^4}\left( {1 + {1 \over {2x}}} \right)-\log {10^3} = \log 10\left( {\root \of (0.75x-0.5) + 2} \right)$

1317003_1008965_ans_21beb17c1af5443e95d641d5ca6bbd35.jpg

Solve $log_{10} tan1^\circ \, + \, log_{10} tan2^\circ \, + \, ........ \, + \, log_{10} tan89^\circ$

$\textbf{Step 1 : Apply the principle properties of logarithm}\\$

$\textbf{Formula Used :}\boldsymbol{\log_{a}{m}+\log_{a}{n}=\log_{a}{m}{n}} ;$ where

$m > 0 , n > 0,a >0,\neq0$

$\textbf{To find : }(log_{10} tan1^\circ \, + \, log_{10} tan2^\circ \, + \, ........ \, + \, log_{10} tan89^\circ)$

$= \, log_{10}(tan1^\circ \, tan2^\circ \, ..... tan44^\circ \, tan45^\circ \, tan46^\circ \, ...........tan89^\circ \, )$

$= \, log_{10}[(tan1^\circ \, tan89^\circ)(tan2^\circ \, tan88^\circ)............(tan44^\circ \, tan46^\circ) \, tan45^\circ]$

$\textbf{Step 2 : Apply the relevant identity/formula}$

$\bigg[\because\tan{1^\circ}\cot{1^\circ}=\tan{2^\circ}\cot{2^\circ}= ... =\tan{1^\circ}\cot{1^\circ}=\tan{44^\circ}\cot{44^\circ}=1\, ; \tan{45^\circ}=1\ \\and\ \tan{(\dfrac{\pi}{2}-\theta})= \cot\theta\bigg]$

$\text{Given expression :}$

$= \, log_{10}[(tan1^\circ \, cot 1^\circ)(tan2^\circ \, cot2^\circ)..........(tan44^\circ \, cot44^\circ) \, tan45^\circ]$

$= \, log_{10} 1 \, = \, 0$

Solve the equations:
$3^x \, 5^y \, = \, 75$
$3^y 5^x \, = \, 45$

Dividing both the equations, we get

$3^{x \, - \, y} \times 5^{y \, - \, x} \, = \, \dfrac{75}{45}$

$\left (\dfrac{3}{5} \right )^{x \, - \, y} \, = \, \left (\dfrac{3}{5} \right )^{- 1}$

Comparing the powers as bases are equal,

$\implies x - y = -1$ .............(1)

Again mulitiplying the equation we get

$(3^x \, \times 3^y ) \, (3^x \, \times 3^y ) \, = 75 \, \times 45$

$15^{x + my } \, = 15^3$

$\implies x + y = 3$ ............(2)

solving (1) and (2) we get ,

$x = 1 , y = 2$

Prove the following identities :
If $a^2 \, + \, b^2 \, = \, 7ab$ , prove that $\log\dfrac{1}{3}(a \, + \, b) \, = \dfrac{1}{2}[\log a +\log b]$

$\textbf{Step 1: Given expression is:}$

$a^2+b^2=7ab$

Adding

$2ab$ on both sides we get,

$a^2+b^2+2ab=9ab$

$\Rightarrow (a+b)^2=9ab\Rightarrow \left(\dfrac{a+b}{3}\right)^2=ab$

$\textbf{Step 2: Now, applying log on both sides we get:}$

$\log \left(\dfrac{a+b}{3}\right)^2=\log (ab)$

$\textbf{Step 3:} \boldsymbol{\log_am^n=n.\log_am}$

$\therefore 2\log\left(\dfrac{a+b}{3}\right)=\log (ab)$

$\textbf{Step 4:} \boldsymbol{\log_am+\log_an=\log_a(mn)}$

$\Rightarrow \log\left(\dfrac{a+b}{3}\right)=\dfrac{\log a+\log b}{2}$

is the required result.

Substituting the values, find $n$
$\cfrac{0.1}{0.2}=\left( \cfrac{10}{80} \right)^{1/n}$ or $\left( \cfrac{1}{2} \right)^{1}=\left( \cfrac{1}{2} \right)^{3/n}$

$\left( \cfrac{1}{2} \right)^{1}=\left( \cfrac{1}{2} \right)^{3/n}$
When bases are same, we can compare powers.
So,

$\cfrac{1}{1}=\cfrac{3}{n} \\ \Rightarrow n=3$

$Express\quad in\quad logarithmic\quad form:\\ { 5 }^{ 3 }=125$

According to the logarithmic form

By definition we know that

$log_ba\leftrightarrow c=b^c=c$

$5^3=125$

$\log_{5}^{125}=3$

The value of x satisfying the equation ${4^{{{\log }_9}^3}} + {9^{{{\log }_2}^4}} = {10^{{{\log }_c}^{83}}}$ is

$4^{\log_{9}^{3}}+9^{\log_{2}^{4}}=10^{\log_{c}^{83}}$

$\Rightarrow (4)^{{1/2}(\log_{3}{3})}+(9)^{2\log_{2}{2}}=10^{\log_{c}(83)}$

$\Rightarrow (4)^{1/2}+(9)^{2}=10^{\log_{c}(83)}$

$\Rightarrow 2+81=10^{\log_{c}(83)}$

$\Rightarrow 83=10^{\log_{c}(83)}$

$\Rightarrow \log_{10}(83)=\log_{c}{(83)}$

$\Rightarrow c=10$

If ${4^{{{\log }_2}2x}} = 36$ , then find x.

$4^{\log_{2}{2x}}=36$ and

$x>0$

$\Rightarrow (2)^{2\log_{2}{2x}}=36$

$\Rightarrow (2)^{\log_{2}{2x}^{2}}=36$

$\Rightarrow (2x)^2=36$

$\Rightarrow (2x)^2-(6)^{2}=0$

$\Rightarrow \left(2x-6\right)\left(2x+6\right)=0$

$\Rightarrow x=3,-3$

$\Rightarrow x\neq -3,\quad x=3$

Solve the equation $\dfrac{3}{2}\log_4(x+2)^2+3=\log_4(4-x)^3+\log_4(6+x)^3$ .

$\dfrac{3}{2}\log_4(x+2)^2+3=\log_4(4-x)^3+\log_4(6+x)^3$

$\dfrac{3}{2}\times 2\log_4(x+2)+3\log_44=3\log_4(4-x)+3\log_4(6+x)$ -----since,

$\log a^n=n\log a$

$3(\log_4(x+2)+\log_44)=3(\log_4(4-x)+\log_4(6+x))$

$\log_4(x+2)+\log_44=\log_4(4-x)+\log_4(6+x)$

$\log_4(x+2)4=\log_4(4-x)(6+x)$ ----- Product law of logarithms

$(x+2)4=(4-x)(6+x)$

$4x+8=-x^2-2x+24$

$x^2+6x-16=0$

$x^2+8x-2x-16=0$

$(x+8)(x-2)=0$

$x=-8$ ,

$x=2$

Find $x$ : $3(2^{x} + 1) -2^{x+2} + 5 =0$

Consider,

$3(2^x+1)-2^{x+2}+5=0$

$3 . 2^x+3-2^2 . 2^x+5=0$ ------Product law

$a^m . a^n=a^{m+n}$

$3 . 2^x+3-4 . 2^x+5=0$

$(3 . 2^x-4 . 2^x)+(5+3)=0$

$-2^x+8=0$

$2^x=8=2^3$

$x=3$ ---base is same, then the powers can be equated

Solve : $\log_3(\sqrt{x}+|\sqrt{x}-1|)=\log_9(4\sqrt{x}-3+4|\sqrt{x}-1|)$

Given

$\log _{ 3 }{ (\sqrt { x } +\left| \sqrt { x } -1 \right| ) } =\log _{ 9 }{ (4\sqrt { x } -3+4\left| \sqrt { x } -1 \right| ) } \\ \Rightarrow \log _{ 3 }{ (\sqrt { x } +\left| \sqrt { x } -1 \right| ) } =\log _{ { 3 }^{ 2 } }{ (4\sqrt { x } -3+4\left| \sqrt { x } -1 \right| ) } \\ \Rightarrow \log _{ 3 }{ (\sqrt { x } +\left| \sqrt { x } -1 \right| ) } =\cfrac { 1 }{ 2 } \log _{ 3 }{ (4\sqrt { x } -3+4\left| \sqrt { x } -1 \right| ) } \quad (\log _{ { a }^{ m } }{ b } =\cfrac { 1 }{ m } \log _{ a }{ b } )\\ \Rightarrow \sqrt { x } +\left| \sqrt { x } -1 \right| =\sqrt { (4\sqrt { x } -3+4\left| \sqrt { x } -1 \right| ) }$

Squaring both sides we get ,

$\Rightarrow x+x+1-2\sqrt { x } +2\sqrt { x } \left| \sqrt { x } -1 \right| =4\sqrt { x } -3+4\left| \sqrt { x } -1 \right| \quad -(i)$

Case1:

$\sqrt { x } -1\ge 0\quad \Rightarrow x\ge 1$

Equation

$(i)$ then reduces to

$2x+1-2\sqrt { x } +2x-2\sqrt { x } =4\sqrt { x } -3+4\sqrt { x } -4\\ \Rightarrow 4x-4\sqrt { x } +1=8\sqrt { x } -7\\ \Rightarrow 4x-12\sqrt { x } +8=0\\ \Rightarrow x-3\sqrt { x } +2=0\\ \Rightarrow x-2\sqrt { x } -\sqrt { x } +2=0\\ \Rightarrow \sqrt { x } (\sqrt { x } -2)-1(\sqrt { x } -2)=0\\ \Rightarrow (\sqrt { x } -1)(\sqrt { x } -2)=0\\ \therefore x=1\quad or\quad x=4$

In the region

$x\ge 1$ ,both

$x=1$ &

$4$ are acceptable.

Case2:

$\sqrt { x } -1<0\quad \Rightarrow x<1$ but

$x>0$ i.e.,

$x\epsilon (0,1)$

Equation

$(i)$ then reduces to

$2x+1-2\sqrt { x } +2\sqrt { x } -2x=4\sqrt { x } -3+4\sqrt { x } \\ 1=1$

which is always true.

$\therefore$ the given equality

$\log _{ 3 }{ (\sqrt { x } +\left| \sqrt { x } -1 \right| ) } =\log _{ 9 }{ (4\sqrt { x } -3+4\left| \sqrt { x } -1 \right| ) }$

holds true for

$x\epsilon (0,1)\cup \{ 1,4\}$

Prove that $x^{\log y-\log z}.x^{\log z-\log x}.x^{\log x-\log y}=1$ .

Now,

$x^{\log y-\log z}.x^{\log z-\log x}.x^{\log x-\log y}$

$=x^{(\log y-\log z)+(\log z-\log x)+(\log x-\log y)}$ [ Using product rule of exponent]

$=x^0$

$=1$ .

The simplified value of $\sqrt {72} + \sqrt {800} - \sqrt {18}$ is

$\sqrt{72}+ \sqrt{800}-\sqrt{18}$

$=\sqrt{3^2\times2^2\times2}+ \sqrt{10^2\times2^2\times2}-\sqrt{3^2\times2}$

$=6\sqrt{2}+ 20\sqrt{2}-3\sqrt{2}$

$=(6+20-3)\sqrt2$

$=23\sqrt2$

Prove that:
$\log(1+2+3)=\log 1+\log 2+\log 3$ .

We have,

$\log(1+2+3)$

$=\log (6)$

$=\log(1.2.3)$

$=\log1+\log 2+\log 3$ .

Find the value of ${49^{\left( {1 - {{\log }_7}2} \right)}} + {5^{ - {{\log }_5}4}}.$

${ 49 }^{ (1-\log _{ 7 }{ 2 } ) }+{ 5 }^{ -\log _{ 5 }{ 4 } }$

${ 49 }^{ \log _{ 7 }{ (\cfrac { 7 }{ 2 } ) } }+{ 5 }^{ \log _{ 5 }{ { (4) }^{ -1 } } }$

${ (\cfrac { 7 }{ 2 } ) }^{ 2 }+{ (4) }^{ -1 }$

$\cfrac { 49 }{ 4 } +\cfrac { 1 }{ 4 } =\cfrac { 50 }{ 4 } =\cfrac { 25 }{ 2 }$

Solve:

$( \dfrac{2 \times 3.0 \times 10^{-25}}{9.1 \times 10^{-31}})^{}$

Solve:

we know,

$\dfrac{a^{m}}{a^{n}}=a^{m-n}$

$( \dfrac{2 \times 3.0 \times 10^{-25}}{9.1 \times 10^{-31}})^{}$

$=\dfrac{2*3.0*10^{6}}{9.1}$

$=0.65934*10^{6}$

$=6.5934*10^{7}$

If $\left(\dfrac{21}{19}\right)^4\times \left(\dfrac{19}{7}\right)^4=b^4$ .Find $b$

$(\frac{21}{19})^4 \times (\frac{19}{7})^4=b^4\\(\frac{21^4}{19^4}) \times (\frac{19^4}{7^4})=b^4\\(\frac{21}{7})^4=b^4\\\therefore b = (\frac{21}{7})=3$

Simplify the following.
${ 3 }^{ 2 }\div { 3 }^{ 4 }$

Let

$= {3^2} \div {3^4}$

Implies that

$=\cfrac{{3 \times 3}}{{3 \times 3 \times 3 \times 3}}$

Implies that

$=\cfrac{1}{{3 \times 3}}$

Hence,

$=\cfrac 19$

Simplify the following.
${ a }^{ x }\div { a }^{ y }$

Let

${a^x} \div {a^y}$

Implies that

$=\cfrac{{{a^x}}}{{{a^y}}}$

Hence,

$={a^{x - y}}$

${\log _{2\sqrt 3 }}1728$

Here we can see that

$2\sqrt{3}$ can be written as

$\sqrt{12}$

Hence,

$\log_{2\sqrt{3}}{1728}=\log_{\sqrt{12}}1728$

$\Rightarrow \log_{2\sqrt{3}}{1728}=\log_{\sqrt{12}}(12)^3$

This can be further simplified as

$\Rightarrow \log_{2\sqrt{3}}{1728}=\log_{\sqrt{12}}(\sqrt{12})^6$

$\Rightarrow \log_{2\sqrt{3}}1728=6\times\log_{\sqrt{12}}\sqrt{12}$

We know that

$\log_{x}x=1$

Therefore,

$\Rightarrow \log_{2\sqrt{3}}1728=6\times 1$

$\Rightarrow \log_{2\sqrt{3}}1728=6$

If $2^{x+1} = 3^{1-x}$ then find the value of x.

$2^{x+1}=3^{-x}$

taking

$\log$ both side

$(x+1\log 2=(1-x)\log 3$

$x=\dfrac{\log 3-\log 2}{\log 1+\log 3}$

Simplify the following.
${ x }^{ 3 }\div { x }^{ 7 }$

Let

${x^3} \div {x^7}$

Implies that

$\cfrac{{{x^3}}}{{{x^7}}}$

Implies that

$={x^{3 - 7}}$

Hence

$={x^{ - 4}}$

using log tables, calculate ${\left[ {\frac{{5.3 \times 2.843}}{{0.80341}}} \right]^{\frac{1}{3}}}$
Using Log tables calculate ${\frac{{{{\left( {4.63} \right)}^2} \times {{\left( {0.08341} \right)}^{\frac{2}{5}}}}}{{{{\left( {0.006743} \right)}^{\frac{1}{4}}}}}}$

$(i)$ Let

$x={\left[\dfrac{5.3\times 2.843}{0.80341}\right]}^{\frac{1}{3}}$

$\Rightarrow\,\log{x}=\log{\left[{\left(\dfrac{5.3\times 2.843}{0.80341}\right)}^{\frac{1}{3}}\right]}$

$\Rightarrow\,\log{x}=\dfrac{1}{3}\log{\left(\dfrac{5.3\times 2.843}{0.80341}\right)}$

$\Rightarrow\,\log{x}=\dfrac{1}{3}\left[\log{5.3}+\log{2.843}-\log{0.80341}\right]$

$=\dfrac{1}{3}\left[0.7242+0.4537+0.0950\right]$

$=\dfrac{1}{3}\times 1.2729=0.4243‬$

$(ii)y=\dfrac{{\left(4.63\right)}^{2}\times{\left(0.08341\right)}^{\frac{2}{5}}}{{\left(0.006743\right)}^{\frac{1}{4}}}$

$\log{y}=\log{\left[\dfrac{{\left(4.63\right)}^{2}\times{\left(0.08341\right)}^{\frac{2}{5}}}{{\left(0.006743\right)}^{\frac{1}{4}}}\right]}$

$=2\log{\left(4.63\right)}+\dfrac{2}{5}\log{\left(0.08341\right)}-\dfrac{1}{4}\log{\left(0.006743\right)}$

$=2\times 0.6655+\dfrac{2}{5}\times -1.0787-\dfrac{1}{4}\times -2.17114$

$=1.331‬-0.43148+0.542785‬=1.442305$

Simplify the following. ${ 3 }^{ 4 }\times { 4 }^{ 4 }$

Let

${3^4} \times {4^4}$

Implies that

${3^4} = 3 \times 3 \times 3 \times 3 = 81$

${4^4} = 4 \times 4 \times 4 \times 4 = 256$

$\left( {{3^4} \times {4^4}} \right) = 81 \times 256$

$=20736$

Simplify the following.
${ 5 }^{ 7 }\div { 5 }^{ 3 }$

Let

${5^7} \div {5^3}$

Implies that

$\cfrac{{5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5}}{{5 \times 5 \times 5}}$

Implies that

$5 \times 5 \times 5 \times 5$

Implies that

$=625$

Simplify: $\dfrac{{16 \times {2^{n + 1}} - 4 \times {2^n}}}{{16 \times {2^{n + 2}} - 2 \times {2^{n + 2}}}}$

The given expression is

$\dfrac{16^{\ast}2^{n+1}-4^{\ast}2^n}{16^{\ast}2^{n+2}-2^{\ast}2^{n+2}}$

$=\dfrac{2^n[16^{\ast}2-4]}{2^{n+2}[16-2]}$

$=\dfrac{2^n{^{\ast}}[32-4]}{2^{n+2}{^{\ast}}14}$

$=\dfrac{28}{2^{n+2-n}{^{\ast}}14}$

$=\dfrac{2}{2^2}=\dfrac{1}{2}$ .

How to find $\log$ of any number?

Always Remember these three properties of logarithm;

$\log(mn)= \log m +\log n$

$\log(m/n)= \log m - \log n$

and $\log(m^n) = n\log m$

Sometimes, When there is no log table around you, the only thing you got is your mind.Here is a method which can be used to calculate the logarithm of a number if there is no log table to use.This is also helpful if the number is too large to be calculated by a calculator.First of all, remember the following basic log values (its no harm to remember these even if you have log table).

$\log2 = 0.301$

$\log3 = 0.477$

$\log5 = 0.699$

$\log7 = 0.845$

these four values can be applied to calculate the log of any number (size does not matter).

and the formula is, $\log(a+b) = \log a + b/(2.42 \times a)$

This is a main method to find any log value.

For example, $\log(737)=?$

$\log(mn)= \log m +\log n$

$=\log (737) = \log (7.37 \times 10^2)$

$=\log (7.37 ) + \log (10^2)$ by using

$=\log (7.37) +2 \quad \quad [ \log 10^n = n]$

$=\log(7 + 0.37) + 2$

Now applying the formula;

$=\log(7)+ 0.37/(2.42\times 7) + 2$

it comes out to be 2.867 which is the log value for $737$ .

now this technique can be applied to find the log of any number.

Solve:
$x=(a^{2/3})^3 \div a^{3/2} \times a^{-3}$ If $a=\dfrac{1}{4}$ . Find $x^{-1}$

We have,

$x=(a^{2/3})^3\div a^{3/2}\times a^{-3}$

$\Rightarrow x= a^2\div a^{-3/2}$

$\Rightarrow x= a^2\times a^{3/2}$

$\Rightarrow x= a^{7/2}$

Put

$a=\dfrac{1}{4}$

Therefore,

$\Rightarrow x= \left(\dfrac{1}{4}\right)^{7/2}$

$\Rightarrow x= \left(\dfrac{1}{2}\right)^7$

$\Rightarrow x= \left(\dfrac{1}{128}\right)$

$\Rightarrow x^{-1}=128$

Hence, this is the answer.

Find $x$ if
a) ${2^x} = 22$
b) ${3^x} = 243$

$(a){2}^{x}=22$

$\Rightarrow\,\log_{2}{{2}^{x}}=\log_{2}{22}$

$\Rightarrow\,x\log_{2}{2}=\log_{2}{2\times 11}$ using

$\log{ab}=\log{a}+\log{b}$

$\Rightarrow\,x\log_{2}{2}=\log_{2}{2}+\log_{2}{11}$

$\Rightarrow\,x=1+\log_{2}{11}$ since

$\log_{a}{a}=1$

$(b){3}^{x}=243$

$\Rightarrow\,\log_{3}{{3}^{x}}=\log_{3}{243}$

$\Rightarrow\,x\log_{3}{3}=\log_{3}{{3}^{5}}$

$\Rightarrow\,x\log_{3}{3}=5\log_{3}{3}$

$\Rightarrow\,x=5$ since

$\log_{a}{a}=1$

Find out the value of n if ${ 3 }^{ n }={ 4 }^{ n-1 }$

${ 3 }^{ x }={ 4 }^{ x-1 }$

$\therefore x=\log _{ 3 }{ { 4 }^{ \left( x-1 \right) } }$

$\therefore x=\left( x-1 \right) \log _{ 3 }{ 4 }$

$\therefore x\left( 1-\log _{ 3 }{ 4 } \right) =-1$

$\therefore x=\cfrac { 1 }{ \log _{ 3 }{ 4 } }$

$x=0.261$

Simplify
${8^{\frac{2}{3}}} - \sqrt 9 \times 10 + \left( {\frac{1}{{144}}} \right)$

${8}^{\frac{2}{3}}-\sqrt{9}\times 10+\dfrac{1}{144}$

$={\left({2}^{3}\right)}^{\frac{2}{3}}-3\times 10+\dfrac{1}{144}$

$={2}^{2}-30+\dfrac{1}{144}$

$=4-30+\dfrac{1}{144}$

$=-26+\dfrac{1}{144}$

$=\dfrac{-75479040+1‬}{144}=\dfrac{75479039‬}{144}$

The value of $2^{\log_{4}{25}}$ is ____

$2^{\log_{4}{25}} = 25^{\log_{4}{2}} = 25^{\cfrac{1}{2}\log_{2}{2}} = 25^{\cfrac{1}{2}} = 5$

Solve for $x$
$\displaystyle {(81)^{\frac{3}{4}}} - {\left( {\frac{1}{{32}}} \right)^{\frac{{ - 2}}{5}}}+ x{\left( {\frac{1}{2}} \right)^{ - 1}}{\times 2^0} = 27$

1321168_1079453_ans_c391ed66311a43e7bab3d0f63055e0dc.jpeg

Solve for $x$
${2^{3x + 3}} = {2^{3x + 1}} + 48.$

1321169_1079456_ans_3c00aab8bb684000b23766e0c87dc039.jpeg

$\log_{e}(101)$ gives $\log_{e}10=2.3026$

Let

$y=\log_e x$

$\Rightarrow \ dy=\dfrac {1}{x}dx$

where

$x=100$ an d

$dx=1$

$\therefore \ dy=0.01$

$\therefore \ y=y+dy$

$=(\log_e 100)+0.01$

$=(2.3026)2+0.01$

$=46052+0.01$

$=4.6152$

If $\log{x}=0$ then $x=?$

1356104_1129350_ans_e930fff5648c42dcb77ea8da7a313463.jpg

$\log\dfrac {20}{10}=?$

$\log \dfrac {20}{10}=\log 2-\log 1=\log 2 \ [\therefore \log 1=0]$

$=0.301$

Evaluate:
$\dfrac{\left(\dfrac{-3}{5}\right)^3\times \left(\dfrac{9}{25}\right)^2\times \left(\dfrac{-18}{125}\right)^0}{\dfrac{-27}{125}\times \left(\dfrac{-3}{5}\right)}$ .

$\cfrac { { (\cfrac { -3 }{ 5 } ) }^{ 3 }\times { (\cfrac { 9 }{ 25 } ) }^{ 2 }\times { (\cfrac { -18 }{ 125 } ) }^{ 0 } }{ (\cfrac { -27 }{ 125 } )\times \cfrac { -3 }{ 5 } } \\ =\cfrac { { (\cfrac { -3 }{ 5 } ) }^{ 3 }\times { (\cfrac { -3 }{ 5 } ) }^{ 4 }\times 1 }{ { (\cfrac { -3 }{ 5 } ) }^{ 3 }\times \cfrac { -3 }{ 5 } } \\ ={ (\cfrac { -3 }{ 5 } ) }^{ 3 }\\ =\cfrac { -27 }{ 125 }$

Solve:
$\log_{x^2-1}(x+1)<1$

given

$\log _{ { x }^{ 2 }-1 }{ \left( x+1 \right) } <1$

Basic conditions :

$x+1> 0\Rightarrow x> -1....(1)$

$({x}^{2}-1)> 0\Rightarrow (x-1)(x+1)> 0$

$x\in (-\infty,-1)\cup (1,\infty)....(2)$

$({x}^{2}-1)\ne 1\Rightarrow x\ne \pm \sqrt{2}...(3)$

Case 1:

$0< {x}^{2}-1< 1$

$\log _{ { x }^{ 2 }-1 }{ \left( x+1 \right) } <1$

$\Rightarrow$

$(x+1)> ({x}^{2}-1)$ (sign changes)

${x}^{2}-x-2< 0$

$(x+1)(x-2)< 0$

$x\in (-1,2)....(4)$

Also

$({x}^{2}-1)< 1\Rightarrow ({x}^{2}-2)< 0$

$(x-\sqrt 2)(x+\sqrt 2)< 0$

$x\in (-\sqrt 2,\sqrt 2).....(5)$

Combining (2), (4), (5)

$x\in (1,\sqrt 2)....(6)$

Case II:

$({x}^{2}-1)> 1\Rightarrow {x}^{2}-2> 0$

$(x-\sqrt 2)(x+\sqrt 2)> 0$

$x\in (-\infty,-\sqrt 2)\cup (\sqrt 2,\infty)....(7)$

$\log _{ { x }^{ 2 }-1 }{ \left( x+1 \right) } <1$

$x+1< {x}^{2}-1$ (sign remains same)

${x}^{2}-x-2> 0$

$(x+1)(x-2)< 0$

$x\in (-\infty,-1)\cup (2,\infty)....(8)$

comining (7) and (8)

$\Rightarrow$

$x\in (-\infty,-\sqrt 2)\cup(2,\infty)....(9)$

combining (1), (2), (3), (6), (9)

$x\in (1,\sqrt 2)\cup (2,\infty)$

$\log_{e}(4.04)$ given $\log_{10}4=0.6021,\log_{10}e=0.4343$

1360979_1129267_ans_bfd163d4646b4fc98dfea960243a91ff.jpg

$\log_{e}(9.01)$ gives $\log_{e}3=1.0986$

Let

$y=\log_e x$

$dy=\dfrac {1}{x}dx$

where

$x=9$ and

$dx=0.01$

$\therefore \ dy=\dfrac {1}{9}\times 0.01$

$\therefore \ y=y+dy$

$=2\log_e 3+0.0011$

$=2\times 1.0986+0.0011$

$=2.1972+0.0011$

$=2.0983$

Find the value of:

$2(6(\sqrt{3})^{5})(\sqrt{3})+20\times (\sqrt{3})^{3}(\sqrt{2})^{3}+6(\sqrt{3})(\sqrt{2})^{5}$

$12.(\surd{3})^{5}.(\surd{3})+20\times (\surd{3})^{3}.(\surd{2})^{3}+6(\surd{3})(\surd{2})^{5}$

$=12.(\surd{3})^{6}+20(\surd{3})^{3}.(\surd{2})^{3}+6(\surd{3})(\surd{2})^{5}$

$=2\surd{2}(6(\surd{3})^{5}+10(\surd{3})^{2}.(\surd{2})^{3}+3(\surd{2})^{5})$

$=2\surd{3}[6.3.3\surd{3}+10.3.2.\surd{2}+3.2.2\surd{2}]$

$=2\surd{3}[54\surd{3}+60\surd{2}+12\surd{2}]$

$=2\surd{3}[54\surd{3}+72\surd{2}]$

$\Rightarrow 108\times 3+144\surd{6}\Rightarrow 324+144\surd{6}$

If we multiply m with ${ \left( \dfrac { 256 }{ 6561 } \right) }^{ -\dfrac { 5 }{ 8 } }$ we get $1$ , then $m =$

According to the question

$m\times\left(\dfrac{256}{6561}\right)^{-\dfrac{5}{8}}=1$

$m\times\left(\dfrac{6561}{256}\right)^{\dfrac{5}{8}}=1$

$m=\dfrac{1}{\left(\dfrac{6561}{256}\right)^{\dfrac{5}{8}}}$

$m=\left(\dfrac{6561}{256}\right)^{\dfrac{5}{8}}$

The value of $\log_{5}\left(\dfrac{1}{625}\right)$ is

$log_5(\dfrac{1}{625})$

$= \dfrac{log_5(1)-log625}{log5}$

$= \dfrac{0-log 5^4}{log^{5}}$

$= \dfrac{-4log5}{log5} = \boxed{-4}$

1146099_1143918_ans_b9d794bff7c64a81844c9b48294693a1.jpg

Compute the following
$\left(\dfrac {1}{49}\right)^{1+\log_{7}2}+5^{-\log_{1/5}7}$

$=\left(\dfrac{1}{49}\right)^{1+log_7^2}+5^{-log_{1/5}^7}$

$=\left(\dfrac{1}{49}\right)^{log_7^{14}}+7^{log_{1/5}^{1/5}}$

$=14^{log_7^{1/49}}+7$

$=(14)^{-2}+7$

$=\left(\dfrac{1}{14}\right)^2+7$ .

1191728_1157724_ans_a05350d0af4c488a805181678a7830ff.jpg

Solve : $2^{3x + 1} + 2^7 = (-4)^4$

$2^{3x+1}+2^7=(-4)^4$

$\implies 2^{3x+1}+2^7=(-1)^4.(2^2)^4$

$\implies 2^{3x+1}+2^7=2^8$ (since,

$(a^m)^n=a^{mn}$ )

$\implies 2^{3x+1}=2^8-2^7$

$\implies 2^{3x+1}=2^7(2-1)$

$\implies 2^{3x+1}=2^7$

when bases are equal then equate the powers

$\implies 3x+1=7$

$\implies 3x=6$

$\implies x=2$

If $\dfrac{{{9^b} \times {3^2} \times {{\left( {{3^{\dfrac{{ - b}}{2}}}} \right)}^{ - 2}} - {{\left( {27} \right)}^b}}}{{{3^{3a}} \times {2^3}}} = \dfrac{1}{{27}}$ , then prove that $a - b = 1$ .

$\dfrac { { 9 }^{ b }\times { 3 }^{ 2 }\times ({ 3 }^{ { \frac { -b }{ 2 } }^{ 2 } })-{ 27 }^{ b } }{ { 3 }^{ 3a }\times { 2 }^{ 3 } } =\dfrac { 1 }{ 27 } \\ \Rightarrow \dfrac { 3^{ 2b }\times { 3 }^{ 2 }\times ({ 3 }^{ { b } })-{ 3 }^{ 3b } }{ { 3 }^{ 3a }\times { 2 }^{ 3 } } =\dfrac { 1 }{ 27 } \\ \Rightarrow \frac { 3^{ 3b }\times { 3 }^{ 2 }-{ 3 }^{ 3b } }{ { 3 }^{ 3a }\times { 2 }^{ 3 } } =\dfrac { 1 }{ { 3 }^{ 3 } } \\ \Rightarrow \dfrac { 3^{ 3b }({ 9 }-{ 1 }) }{ { 3 }^{ 3a }\times { 8 } } =\dfrac { 1 }{ { 3 }^{ 3 } } \\ \Rightarrow \frac { 3^{ 3b } }{ { 3 }^{ 3a } } =\dfrac { 1 }{ { 3 }^{ 3 } } \\ \Rightarrow { 3b }-{ 3a }={ -3 }\\ { a }-{ b }={ 1 }\\$

$-9\log_{3}^{(b + 2)} = ?$ .

$-9\log_3^{b+2}=18$

$\log_3^{b+2}=-2$

$\implies b+2=3^{-2}$

$\implies b=\cfrac{1}{9}-2=\cfrac{-17}{9}$

Find the value of
$\log_{2} { 5^{ \frac{1}{3} } }$

$\log_2(5)^{1/3}=\dfrac{1}{3}\log_2(5)\Rightarrow \dfrac{1}{3}\dfrac{\log_{10}5}{\log_{10}2}$

$=\dfrac{1}{3}\times \dfrac{0.7}{0.3}=\dfrac{7}{9}=0.717$

Expand $\dfrac{5x^2+4x+7}{\left(2x+3\right)^{\frac{3}{2}}}$

Given

$\dfrac{5x^2+4x+7}{(2x+3)^{\frac{3}{2}}}$

Applying exponent rule

$(a)^{b+c}=a^b.a^c$

$(2x+3)^{\frac{3}{2}}=(2x+3)^{1+\frac{1}{2}}=(2x+3)\sqrt{2x+3}$

$=\dfrac{5x^2+4x+7}{(2x+3)\sqrt{2x+3}}$

Solve:
$8x^7+x^7+x=?$
if $x=7$

${ 8x }^{ 7 }+{ x }^{ 7 }+x={ 9x }^{ 7 }+x$

Bow putting

$x=7$ we have

${ 9\left( 7 \right) }^{ 7 }+7=7411887+7=74,11,894$

For any base show that
log (1+2+3)= log 1+ log 2+ log3.
Note that , in general,
log (a+b+c) $\neq$ log a+ log b +log c.

To prove:-

$\log{\left( 1 + 2 + 3 \right)} = \log{1} + \log{2} + \log{3}$

Taking L.H.S.-

$\log{\left( 1 + 2 + 3 \right)}$

$= \log{6}$

$= \log{\left( 1 \times 2 \times 3 \right)}$

$= \log{1} + \log{2} + \log{3} \; \left( \because \log{ab} = \log{a} + \log{b} \right)$

$=$ R.H.S.

Hence proved.

Solve the following:-
$\left(\dfrac{2^6 \times 3^4}{6^3}\right)^2\times \left(\dfrac{3^3 \times 2^5}{3 \times 8}\right)^3$

$\left(\dfrac{2^6 \times 3^4}{6^3}\right)^2\times \left(\dfrac{3^3 \times 2^5}{3 \times 8}\right)^3$

Let

$A= (\dfrac{2^{3}\times 3^{3}\times 2^{3}\times 3}{6^{3}})^{2} \times (\dfrac{3^{2}\times 2^{3}\times 2^{2}}{8})^{3}$

$= (8\times 3)^{2}\times (9 \times 4)^{3}$

$= 24^{2}\times 36^{3}$

$A= 26873856.$

$\left(\dfrac{2^6 \times 3^4}{6^3}\right)^2\times \left(\dfrac{3^3 \times 2^5}{3 \times 8}\right)^3$

$= 26873856.$

$log_x\ a\ +- log_{ax}a\ +\ 3log_{a^{2}x}\ a\ =\ 0$

$\log_{x}{a} + \left( -\log_{ax}{a} \right) + 3 \log_{{a}^{2}x}{a} = 0$

$\log_{x}{a} - \log_{ax}{a} + 3 \log_{{a}^{2}x}{a} = 0$

$\cfrac{1}{\log_{a}{x}} - \cfrac{1}{\log_{a}{ax}} + \cfrac{3}{\log_{a}{{a}^{2}x}} = 0$

$\cfrac{1}{\log_{a}{x}} - \cfrac{1}{\log_{a}{a} + \log_{a}{x}} + \cfrac{3}{\log_{a}{{a}^{2}} + \log_{a}{x}} = 0$

$\cfrac{1}{\log_{a}{x}} - \cfrac{1}{1 + \log_{a}{x}} + \cfrac{3}{2 + \log_{a}{x}} = 0$

Let

$y = \log_{a}{x}$

$\therefore \cfrac{1}{y} - \cfrac{1}{1 + y} + \cfrac{3}{2 + y} = 0$

$\Rightarrow \cfrac{\left( 1 + y \right) \left( 2 + y \right) - y \left( 2 + y \right) + 3y \left( 1 + y \right)}{y \left( 1 + y \right) \left( 2 + y \right)} = 0$

$\Rightarrow \left( 1 + y \right) \left( 2 + y \right) - y \left( 2 + y \right) + 3y \left( 1 + y \right) = 0$

$\Rightarrow \left( 2 + y + 2y + {y}^{2} \right) - \left( 2y + {y}^{2} \right) + \left( 3y + 3{y}^{2} \right) = 0$

$\Rightarrow 2 + 3y + {y}^{2} - 2y - {y}^{2} + 3y + 3 {y}^{2} = 0$

$\Rightarrow 3{y}^{2} + 4y + 2 = 0$

Now by quadratic formula, we have

$y = \cfrac{-4 \pm \sqrt{16 - 24}}{6}$

$\Rightarrow y = \cfrac{-2 \pm 2 \sqrt{2} i}{3}$

Therefore,

$\log_{a}{x} = \cfrac{-2 \pm 2 \sqrt{2} i}{3}$

$\Rightarrow x = a^ {{\frac{-2 \pm 2 \sqrt{2} i}{3}}}$

Hence

$x = {a}^{\frac{-2 \pm 2 \sqrt{2} i}{3}}$ .

Solve and write the answer in one term.
$\log 2 + 1$ .

Solution-

$\log 2+1$

Suppose

$\log$ is on base

$10$

$\therefore \log 10^2+1$

$1$ can be written as

$\log_{10}10$ .

$=\log _{10^2}+\log_{10}10$

$=\log_{10}{(2\times 10)}$

$(\log a+\log b=\log ab)$ .

$=\log_{10}(20)$

$=\log_{10}{20}$

If $\log 2=0.3010,\log 3=0.4771,\log 7=0.8451$ and $\log 11=1.0414$ , then find the value of the following :
$\log \left(\dfrac {11}{7}\right)^{5}$

$\log \left (\dfrac{11}{7} \right )^5$

$=5\log \left (\dfrac{11}{7} \right )$

$=5(\log 11 - \log 7)$

$=5(1.0414-0.8451)$

$=0.9815$

If $log _{2\sqrt 3} 1728=x$ then find x

$log_4m=1.5$
so

$m=(4)^{1.5}$

$m=(40^{3/2}$

$m=(z)^{2x3/2}$

$m=2^3$

$m=8$
by using property log

$log a_b=C$

$b=a^c$

Solve:
$\log_{3}{\left(\dfrac{1}{27}\right)}=-3$

1193563_1223871_ans_d15881d11676450a9eba70f74465b2f8.jpeg

Evaluate $\log _ { 3 } \left( \log _ { 9 } x + \frac { 1 } { 2 } + 9 ^ { x } \right) = 2 x$

$\log_3\left[ \log_9 x+\dfrac{1}{2}+9^x\right] =2x$

$\Rightarrow \log_9 x+\dfrac{1}{2}+3^{2x}=3^{2x}$ ( using property of

$\log$ )

$\Rightarrow \log_9 x+\dfrac{1}{2}=0$

$\log_9 x=\dfrac{-1}{2}$

$x=9^{-1/2}$

$x=\dfrac{1}{9^{1/2}}$

$\therefore x=\dfrac{1}{3}$

Hence, the answer is

$\dfrac{1}{3}.$

Solve:
$\left(-\dfrac {1}{2}\right)^{3}\times \left(-\dfrac {1}{2}\right)^{4}$

$\left( \dfrac{-1}{2}\right)^3\times \left( \dfrac{-1}{2}\right)^4$

$=\dfrac{-1}{8}\times \dfrac{1}{16}$

$=\dfrac{-1}{128}$

Hence, the answer is

$\dfrac{-1}{128}.$

Find the value of $2[(2^{-1}\times 4^{-1})\div 2^{-2}]$ .

Given,

$2[(2^{-1} \times 4^{-1}) \div 2^{-2})$

$2[(\frac{1}{2} \times \frac{1}{4}) \div \frac{1}{4}]$

$2[(\frac{1}{8}) \div \frac{1}{4}]$

$2[(\frac{1}{8}) \times \frac{4}{1}]$

$2(\frac{1}{2})$

$1$

If $\log 2=0.3010,\log 3=0.4771,\log 7=0.8451$ and $\log 11=1.0414$ , then find the value of the following :
$\log 36$

$\log 36$

$=\log 6^2$

$=2\log 6$

$=2\log (2\times 3)$

$=2\log 2 +2\log 3$

$=2(0.3010)+2(0.4771)$

$=1.5562$

${(x \div y)^{ - 1}} = {x^{ - 1}} \div {y^{ - 1}}$ by taking $x = \frac{7}{{11}}$

Consider the given equation,

${{\left( x\div y \right)}^{-1}}={{x}^{-1}}\div {{y}^{-1}}$ by taking $x=\dfrac{7}{11}$

${{\left( x\div y \right)}^{-1}}={{x}^{-1}}\div {{y}^{-1}}$

$\dfrac{1}{\left( x\div y \right)}=\dfrac{1}{x}+\dfrac{1}{y}$

$\dfrac{1}{\dfrac{x}{y}}=\dfrac{1}{x}+\dfrac{1}{y}$

$\dfrac{y}{x}=\dfrac{1}{x}+\dfrac{1}{y}$

$\dfrac{y}{\dfrac{7}{11}}=\dfrac{1}{\dfrac{7}{11}}+\dfrac{1}{y}$

$y=1+\dfrac{7}{11}.\dfrac{1}{y}$

$11{{y}^{2}}-11y-7=0$

$y=\dfrac{11\pm \sqrt{121-4\times 11\times \left( -7 \right)}}{2\times 11}$

$y=\dfrac{11\pm \sqrt{429}}{4}$

$y=\dfrac{11+\sqrt{429}}{4}\,or\,y=\dfrac{11-\sqrt{429}}{4}$

Hence, this is the answer.

If $\log 2=0.3010,\log 3=0.4771,\log 7=0.8451$ and $\log 11=1.0414$ , then find the value of the following :
$\log \dfrac {42}{11}$

$\log \dfrac{42}{11}$

$=\log 42 - \log 11$

$=\log (7\times 2 \times 3) - \log 11$

$=\log 7 +\log 2 + \log 3 - \log 11$

$=0.8451+0.3010+0.4771-1.0414$

$=0.5818$

Solve :- ${{{5^6}} \over {{5^3} \times {5^3}}}$

1142411_1224852_ans_7e98fa9cb2f04ee0b42a286ef68abbd7.jpg

If $\log 2=0.3010,\log 3=0.4771,\log 7=0.8451$ and $\log 11=1.0414$ , then find the value of the following :
$\log 5^{1/3}$

$\log 5^{\frac{1}{3}}$

$=\dfrac{1}{3} \log 5$

$=\dfrac{1}{3} \log\left ( \dfrac{10}{2} \right )$

$=\dfrac{1}{3} (\log 10 -\log 2)$

$=\dfrac{1}{3} (1-0.3010)$

$=0.233$

If $\log 2=0.3010,\log 3=0.4771,\log 7=0.8451$ and $\log 11=1.0414$ , then find the value of the following :
$\log 70$

$\log 70$

$=\log (7\times 10)$

$=\log 7 + \log 10$

$=0.8451+1$

$=1.8451$

${\log _2}x + \frac{1}{2}{\log _2}(x + 2) = 2$

$\log2^x+1/2 \log_{2}(x+2)=2$

$\log _{ 2 }{ x } +\log _{ 2 }{ \left( x+2 \right)^{ { y }_{ 2 }} } =\log _{ 2 }{ { 2 }^{ 2 } }$

$\log_{2}[x(x+2)^{y_{2}}]=\log_{2}4$

$x(x+2)^{y_{2}}=4$

$(x+2)^{y_{2}}=4/x$ [sequencing both sides]

$x+2=\dfrac {16}{x^2}$

$x^{2}(x+2)=16$

$x^{3}+2x^{2}-16=0$

$(x-2)(x^{2}+4x+8)=0$

$x=2$

1347512_1232750_ans_3b6e65a30fc442d6a2f745e0b205127e.PNG

Find the value of $x$ if $\displaystyle 14 \times {10^{ - 4}} = \log {{500} \over x}$

Given,

$\Rightarrow 14\times 10^{-4}=\log\dfrac{500}{x}$

$\Rightarrow 14\times 10^{-4}=\log_{10}500-\log_{10}x \quad \dots \left(\log \left(\dfrac{a}{b}=\log a-\log b\right)\right)$

$\Rightarrow 14\times 10^{-4}=\log_{10}(5\times 100)-\log_{10}x$

$\Rightarrow 14\times 10^{-4}=\log_{10}5+\log_{10}{10^2}-\log_{10}x$

$\Rightarrow 14\times 10^{-4}=\log_{10}5+2\log_{10}{10}-\log_{10}x \quad \dots (n \log_ab=\log_ab^n )$

$\Rightarrow \log_{10}x=\log_{10}5+2-\dfrac{14}{10^4}$

$\Rightarrow x=10^{\log_{10} 5}+10^2-10^{\frac{14}{104}}$

$\Rightarrow x=5+100-10^{14\times 10^{-4}}$

$\Rightarrow x=105-10^{14\times 10^{-4}}$ .

$4^{3.5} : 2 ^5$ is the same as

We have,

$4^{3.5}:2^5$

$=(2^2)^{3.5}:2^5$

$=2^7:2^5$

$=\dfrac{2^7}{2^5}$

$=2^2$

$=4$

Hence, this is the answer.

If $\log 2=0.3010,\log 3=0.4771,\log 7=0.8451$ and $\log 11=1.0414$ then find the value of the following
$\log \dfrac {121}{120}$

$\log \dfrac{121}{120}$

$=\log 121 -\log 120$

$=2\log 11 -\log (3\times 2^3\times 5)$

$=2\log 11 -\log 3 - 3\log 2 -\log 5$

$=2\log 11 -\log 3 - 3\log 2 -\log \left ( \dfrac{10}{2} \right )$

$=2\log 11 -\log 3 - 3\log 2 -(\log 10 -\log 2)$

$=2\log 11 -\log 3 - 3\log 2 -1+\log 2$

$=2\log 11 -\log 3 - 2\log 2 -1$

$=2(1.0414)-0.4771-2\times 0.3010-1$

$=3.604\times 10^{-3}$

${2}^{9}\times{2}^{5}$

$2^9\times 2^5$

$2^5\times 2^4\times 2^5$

$2^2\times 2^2\times 2\times 2^2\times 2^2\times 2^2\times 2^2\times 2$

$4\times 4\times 2\times 4\times 4\times 4\times 4\times 2$

$=16384$ .

Find the value of $\log 7$ using series expansion.

We have,

Value of $\log 7=?$

Now,

We know that,

$\log 2=0.3010$

$\log 3=0.4771$

Therefore,

$\log 7=\log 2+\log 2+\log 3$

$\log 7=0.3010+0.3010+0.4771$

$\log 7=1.07918$

Hence, this is the answer.

Find the value of $\log_2 2^{2^2}$ .

Now,

Let

$a=\log_2 2^{2^2}$

$=\log_2 2^{4}$

$=4\log_2 2 \quad \dots ( n \log_ab=\log_ab^n)$

$a=4$ .

$\log_2 2^{2^2}=4$

The value of a for which $\log _ { a } ( 1 \times 2 ) - \log _ { a } ( 2 \times 3 ) + \log _ { a } ( 3 \times 4 ) - \log _ { a } ( 4 \times 5 ) + \log _ { a } ( 5 \times 6 ) = \log _ { e } 36$ is ...

1141785_1248878_ans_3c36fe6d89664f2db0f8fb8684e75edb.jpg

Simplify the following using laws of exponent
$(\dfrac{3x^2y^2}{axy^4})^2$

We have,

$\left [\dfrac{3x^2y^2}{axy^4} \right ]^2$

$=\left [\dfrac{9x^4y^4}{a^2x^2y^8} \right ]$

$=\left [\dfrac{9x^2}{a^2y^4} \right ]$

Hence, this is the answer.

Simplify the following using laws of exponent
$[7^8 \times 12^8]\div [14^8 \times 6^8]$

We have,

$[7^8\times 12^8]\div[14^8\times 6^8]$

$=\dfrac{7^8\times 12^8}{14^8\times 6^8}$

$=\dfrac{7^8\times (2\times 6)^8}{(2\times 7)^8\times 6^8}$

$=\dfrac{7^8\times 2^8 \times 6^8}{2^8 \times 7^8\times 6^8}$

$=1$

Hence, this is the answer.

Using $\log{2}=0.3010$ and $\log{3}=0.4771$ and $\log{7}=0.8451$ find the value of $\log{6}^{50}$

$\log{{6}^{50}}=50\log{6}$

$=50\left(\log{2}+\log{3}\right)$

$=50\left(0.3010+0.4771\right)$

$=50\times 0.7781=38.905$

Find $n$
$\dfrac {9^{n+1}+4^{n+1}}{9^{n}+4^{n}}=6$

We have,

$\begin{array}{l} \dfrac { { { 9^{ n+1 } }+{ 4^{ n+1 } } } }{ { { 9^{ n } }+{ 4^{ n } } } } \\ { 9^{ n } }\cdot 9+{ 4^{ n } }\cdot 4=6\left( { { 9^{ n } }+{ 4^{ n } } } \right) \\ 9\cdot { 9^{ n } }+{ 4^{ n } }\cdot 4-6\times { 9^{ n } }-6\times { 4^{ n } }=0 \\ { 9^{ n } }\left( { 9-6 } \right) +{ 4^{ n } }\left( { 4-6 } \right) =0 \\ 3\cdot { 9^{ n } }-2\cdot { 4^{ n } }=0 \\ 3\cdot { 9^{ n } }=2\cdot { 4^{ n } } \\ \dfrac { { { 9^{ n } } } }{ { { 4^{ n } } } } =\dfrac { 2 }{ 3 } \\ { \left( { \dfrac { 3 }{ 2 } } \right) ^{ 2n } }={ \left( { \dfrac { 3 }{ 2 } } \right) ^{ -1 } } \\ 2n=-1 \\ n=\dfrac { { -1 } }{ 2 } \end{array}$

Hence, this is the answer.

Simplify and express:
$\displaystyle \frac{2^{3}\times 3^{4}\times 4}{3\times 32}$

We have,

$= \dfrac{{{2^3} \times {3^4} \times 4}}{{3 \times 32}}$

$= \dfrac{{2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 4}}{{3 \times 32}}$

$= \dfrac{{3 \times 3 \times 3 \times 3}}{3}$

$= 27$

Then,

We get

$=27$

Solve:
$2^{5}\times 2^{4}\div 2^{3}$

We have,

${2^5} \times {2^4} \div {2^3}$

$= \dfrac{{{2^5} \times {2^4}}}{{{2^3}}}$

$= \dfrac{{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}}{{2 \times 2 \times 2}}$

Hence,

We get

$= 64$

Prove that
$\dfrac { 1 }{ { log }_{ ab }(abc) } +\dfrac { 1 }{ { log }_{ bc }(abc) } +\dfrac { 1 }{ { log }_{ ca }(abc) } =2$

$\dfrac{1}{log_{ab}(abc)}+\dfrac{1}{log_{bc}(abc)}+\dfrac{1}{log_{ca}(abc)}$

$\Rightarrow \dfrac{log ab}{log abc}+\dfrac{log bc}{log abc}+\dfrac{log ca}{log abc}$

$\left[log_ba=\dfrac{log_ca}{log_cb}\right]$

$\Rightarrow \dfrac{log ab+log bc+log ca}{log abc}$

$\Rightarrow \dfrac{log a+log b+log b+log c+log c+log a}{log abc} \quad \dots ( \log a+\log b=\log(ab))$

$\Rightarrow \dfrac{log(abc)^2}{log(abc)}$

$\Rightarrow \dfrac{2log abc}{log abc} \quad \dots ( n \log_ab=\log_ab^n )$

$\Rightarrow 2$

$\therefore \dfrac{1}{log_{ab}(abc)}+\dfrac{1}{log_{bc}(abc)}+\dfrac{1}{log_{ca}(abc)}=2$

Hence, proved.

If ${ a }^{ 2 }+{ b }^{ 2 }=3ab,$ show that $\log\left( \dfrac { a+b }{ \sqrt { 5 } } \right) =\dfrac { 1 }{ 2 } \left( \log a+\log b \right)$

We have,

${{a}^{2}}+{{b}^{2}}=3ab$

On adding $2ab$ both side and we get,

${{a}^{2}}+{{b}^{2}}+2ab=3ab+2ab$

$\Rightarrow {{\left( a+b \right)}^{2}}=5ab$

On taking $\log$ both side

$\dfrac{\left( a+b \right)}{\sqrt{5}}=\sqrt{ab}$

$\Rightarrow \log \dfrac{\left( a+b \right)}{\sqrt{5}}=\log \sqrt{ab}$

$\Rightarrow \log \dfrac{\left( a+b \right)}{\sqrt{5}}=\log \sqrt{a}+\log \sqrt{b}$

$\Rightarrow \log \dfrac{\left( a+b \right)}{\sqrt{5}}=\log {{\left( a \right)}^{\dfrac{1}{2}}}+\log {{\left( b \right)}^{\dfrac{1}{2}}}$

$\Rightarrow \log \dfrac{\left( a+b \right)}{\sqrt{5}}=\dfrac{1}{2}\operatorname{loga}+\dfrac{1}{2}\log b$

$\Rightarrow \log \dfrac{\left( a+b \right)}{\sqrt{5}}=\dfrac{1}{2}\left( \operatorname{loga}+\log b \right)$

Hence, this is the answer.

$32^{1/5}$ is equal to ___.

We've,

$32^{\frac{1}{5}}$

$=(2^5)^{\frac{1}{5}}$

$=2$ .

solve : $\left[ \left\{ \left( \dfrac { - 1 } { 3 } \right) ^ { 2 } \right\} ^ { - 2 } \right] ^ { - 1 }$

$\begin{array}{l} We\, have \\ { \left[ { { { \left\{ { { { \left( { \frac { { -1 } }{ 3 } } \right) }^{ 2 } } } \right\} }^{ -2 } } } \right] ^{ -1 } } \\ ={ \left[ { { { \left\{ { \frac { 1 }{ 9 } } \right\} }^{ -2 } } } \right] ^{ -1 } }\, \, \, \, \, \left[ { { { \left( { \frac { a }{ b } } \right) }^{ -1 } }=\frac { b }{ a } } \right] \\ ={ \left[ { { { \left\{ { \frac { 9 }{ 1 } } \right\} }^{ 2 } } } \right] ^{ -1 } } \\ ={ \left[ { 81 } \right] ^{ -1 } } \\ =\frac { 1 }{ { 81 } } \\ It\, is\, the\, required\, answer. \end{array}$

Find the value of $x$ if ${25}^{x}\sqrt[x]{{5}^{x-1}}=1$

${25}^{x}\sqrt[x]{{5}^{x-1}}=1$

$\Rightarrow {5}^{2x}{\left({5}^{x-1}\right)}^{\frac{1}{x}}=1$

$\Rightarrow {5}^{2x}\left({5}^{1-\frac{1}{x}}\right)=1$

$\Rightarrow {5}^{2x+1-\frac{1}{x}}={5}^{0}$

Since bases are the same we can equate the powers.

$2x+1-\dfrac{1}{x}=0$

$\Rightarrow 2{x}^{2}+x-1=0$

$\Rightarrow 2{x}^{2}+2x-x-1=0$

$\Rightarrow 2x\left(x+1\right)-1\left(x+1\right)=0$

$\Rightarrow \left(x+1\right)\left(2x-1\right)=0$

$\therefore x=-1,\,\dfrac{1}{2}$

${ \left[ { \left( \dfrac { 5 }{ 7 } \right) }^{ 2 } \right] }^{ 3 }\times { 7 }^{ 6 }\div { \left( { 5 }^{ 3 } \right) }^{ 2 }$ is equal to ____.

$\left [\left(\frac{5}{7} \right )^2 \right ]^3\times 7^{6}\ \div (5^{3})^2$

$\left(\frac{5}{7} \right )^6 \times 7^{6}\ \div (5^{3})^2$

$\left(\frac{5^6}{7^6} \right ) \times 7^{6}\ \div (5^{6})$

$5^6\div 5^6$

$5^0$

$1$ (

$as\ \ \ a^0 = 1)$

${ log }_{ 10 }2+{ 16log }_{ 10 }(\frac { 16 }{ 15 } )+{ 12log }_{ 10 }(\frac { 25 }{ 24 } )+{ 7log }_{ 10 }(\frac { 81 }{ 80 } )$

$\log_{10}2+16\log_{10}+12\log_{10}(\cfrac{25}{24})+7\log_{10}(\cfrac{81}{80})$

$=\log_{10}(2\times (\cfrac{16}{15})^{16}\times(\cfrac{25}{24})^{12}\times(\cfrac{81}{80})^7)$

$=\log_{10}(2\times3^0\times5)$

$=\log_{10}10=1$

Express $\log _{ 10 }{ 2 } +1\quad in\quad the\quad from\quad of\quad \log _{ 10 }{ x. }$

$\log_{10}{2}+1$

$=\log_{10}{2}+\log_{10}{10}$ since

$\log_{a}{a}=1$

$=\log_{10}{20}$ since

$\log_{a}{b}+\log_{a}{c}=\log_{a}{bc}$

is of the form

$\log_{10}{x}$ where

$x=20$

Find the value of :
(i) $\log_{1/2}{8}$
(ii) $\log_5{0.008}$
(iii) $\log_5{3125}$
(iv) $\log_7{\sqrt[3]{7}}$

(i)

${ \log }_{ \frac { 1 }{ 2 } }8={ \log }_{ \frac { 1 }{ 2 } }{ \left( \dfrac { 1 }{ 2 } \right) }^{ -3 }=-3$

(ii)

${ \log }_{ 5 }0.008={ \log }_{ 5 }\dfrac { 1 }{ 125 } ={ \log }_{ 5 }{ \left( 5 \right) }^{ -3 }=-3$

(iii)

${ \log }_{ 5 }3125={ \log }_{ 5 }{ \left( 5 \right) }^{ 5 }=5$

(iv)

${ \log }_{ 7 }\sqrt [ 3 ]{ 7 } ={ \log }_{ 7 }{ \left( 7 \right) }^{ 1/3 }=1/3$

Hence, this is the answer.

The value of $log\, 0.008........$

$log_{10}0.008$

$\Rightarrow log_{10}8\times 10^{-3}$ [ log ab = log a+ log b]

$\Rightarrow log_{10}8+log_{10}10^{-3}$

$\Rightarrow log_{10}2^{3}+(-3) log_{10}10$

$\Rightarrow 3log_{10}2+(-3)$ [ log ab = b log a]

$\Rightarrow 3(0.301)-3$

$\Rightarrow -2.097$

1184221_1364852_ans_8bf4b26ab2e84030879826ee9b18d399.jpg

Simplify: $\sqrt [5] {64}$

$\sqrt[5]{64}=\sqrt[5]{2\times 2\times 2\times 2\times 2\times 2}=\sqrt[25]{2}$ .
1189560_1356053_ans_7e013026009549c9866f2e8233af4449.jpg

Find the approximate value of
(i) ${ log }_{ e }101$ , given that ${ log }_{ e }10=2.3026$
(ii) ${ 3 }^{ 2.04 }$ given that ${ log }_{ e }3=1.0986$ .

1222653_1322509_ans_3ea85be25804432e85969909522afe7a.jpg

If $a=log_{24}12, b=log_{36}24, c=log_{48}36$ . Prove that $1+abc=2bc$ .

$a=\log ^{12}_{24}$

$b=\log ^{24}_{36}$

$c=\log^{36}_{48}$

$abc=\log^{12}_{24}\log^{24}_{36} \log^{36}_{48}$

$=\cfrac {\log 12}{\log 24}\cfrac {\log 24}{\log 36} \cfrac {\log 36}{\log 48}$

$ab=\log^{12}_{48}$

$1+abc=\log^{48}_{48}+\log^{12}_{48}$

$\log ab= \log a+\log b$

$=\log ^{48 \times 12}_{48}=\log ^{12 \times 12 \times 4}_{48}=\log ^{24^2}_{48}=2 \log ^{24}_{48}$

$2bc=2 \times \log^{24}_{36}\times \log^{36}_{48}$

$=2 \log ^{24}_{48}$

$\therefore 1+abc=2bc$ .

$\log _33+\log _416$

$\log _33+\log _416\\1+\log_44^2\\1+2\log_44=1+2=3$

Write ${2}^{10}=1024$ in logarithmic form.

${2}^{10}=1024$

Applying

$\log$ both sides, we get

$\log_{2}{{2}^{10}}=\log_{2}{1024}$

$\Rightarrow 10\log_{2}{2}=\log_{2}{1024}$

using

$\log{{a}^{m}}=m\log{a}$

$\Rightarrow 10=\log_{2}{1024}$ using

$\log_{a}{a}=1$

$\therefore \,\log_{2}{1024}=10$

$\log _6216+\log _7343$

$\log _6216+\log _7343\\\log _66^3+\log_77^3\\3\log_66+3\log _77\\3+3=6$

Find the value of $\log_{\sqrt{2}} 16$ .

Now,

$\log_{\sqrt{2}} 16$

$=\log_{\sqrt{2}} 2^4$

$=\log_{\sqrt{2}}(\sqrt{2})^8$

$=8\log_{\sqrt{2}}\sqrt{2}$

$=8$ .

Write ${3}^{5}=243$ in logarithmic form.

${3}^{5}=243$

Applying

$\log$ both sides, we get

$\log_{3}{{3}^{5}}=\log_{3}{243}$

$\Rightarrow 5\log_{3}{3}=\log_{3}{243}$ (using

$\log{{a}^{m}}=m\log{a}$ )

$\Rightarrow 5=\log_{3}{243}$ (using

$\log_{a}{a}=1$ )

$\therefore \,\log_{3}{243}=5$

Find the value of
i) ${ 2 }^{ 6 }$
ii) ${ 9 }^{ 3 }$
iii) ${ 11 }^{ 2 }$

Now,

${ 2 }^{ 6 }=2 \times 2 \times 2 \times 2\times 2 \times 2=64$

ii)

${ 9 }^{ 3 }=9 \times 9 \times 9=729$

iii)

${ 11 }^{ 2 }11 \times 11=121$ .

Simplify: $\dfrac {a^{m}\times a^{n}}{a^{m-n}}$

Now,

$\dfrac {a^{m}\times a^{n}}{a^{m-n}}$

$=\dfrac {a^{m+n}}{a^{m-n}}$

$=a^{m+n-m+n}$

$=a^{2n}$ .

find the value of $\dfrac { \log _{ 4 }{ 7 } }{ \log _{ 4 }{ 5 } } -\dfrac { \log _{ 9 }{ 5 } }{ \log _{ 9 }{ 7 } }$

$\dfrac { \log _{ 4 }{ 7 } }{ \log _{ 4 }{ 5 } } -\dfrac { \log _{ 9 }{ 5 } }{ \log _{ 9 }{ 7 } }$

We know that

$\dfrac{\log_{a}{b}}{\log_{a}{c}}=\log_{c}{b}$

$\therefore \dfrac{\log_{4}{7}}{\log_{4}{5}}-\dfrac{\log_{9}{5}}{\log_{9}{7}}$

$=\log_{5}{7}-\log_{7}{5}$

We know that

$\log_{b}{a}=\dfrac{1}{\log_{a}{b}}$

$\dfrac{\log_{4}{7}}{\log_{4}{5}}-\dfrac{\log_{9}{5}}{\log_{9}{7}}=\log_{5}{7}-\dfrac{1}{\log_{5}{7}}$

$=\dfrac{{\left(\log_{5}{7}\right)}^{2}-1}{\log_{5}{7}}$

Take

$1={\left(\log_{5}{5}\right)}^{2}$

$=\dfrac{{\left(\log_{5}{7}\right)}^{2}-{\left(\log_{5}{5}\right)}^{2}}{\log_{5}{7}}$

$=\dfrac{\left(\log_{5}{7}+\log_{5}{5}\right)\left(\log_{5}{7}-\log_{5}{5}\right)}{\log_{5}{7}}$

$=\dfrac{\left(\log_{5}{35}\right)\left(\log_{5}{\dfrac{7}{5}}\right)}{\log_{5}{7}}$

$\dfrac{\log_{4}{7}}{\log_{4}{5}}-\dfrac{\log_{9}{5}}{\log_{9}{7}}=\log_{7}{35}\log_{5}{\dfrac{7}{5}}$

Show that $\log{\left(\dfrac{108}{605}\right)}=2\log{2}+3\log{3}-\log{5}-2\log{11}$

L.H.S

$=\log{\left(\dfrac{108}{605}\right)}$

$=\log{\left(\dfrac{2\times 2\times 3\times 3\times 3}{5\times 11\times 11}\right)}$

$=\log{\left(\dfrac{{2}^{2}\times {3}^{3}}{5{11}^{2}}\right)}$

We know that

$\log{\left(\dfrac{a}{b}\right)}=\log{a}-\log{b}$

$=\log{{2}^{2}}+\log{{3}^{3}}-\log{5}-\log{{11}^{2}}$

We know that

$\log{{a}^{m}}=m\log{a}$

$=2\log{2}+3\log{3}-\log{5}-2\log{11}$

$=$ R.H.S

$\therefore \log{\left(\dfrac{108}{605}\right)}=2\log{2}+3\log{3}-\log{5}-2\log{11}$

Hence, proved.

Write ${6}^{3}=216$ in logarithmic form.

${6}^{3}=216$

Applying

$\log$ both sides, we get

$\log_{6}{{6}^{3}}=\log_{6}{216}$

$\Rightarrow 3\log_{6}{6}=\log_{6}{216}$ (using

$\log{{a}^{m}}=m\log{a}$ )

$\Rightarrow 3=\log_{6}{216}$ (using

$\log_{a}{a}=1$ )

$\therefore \,\log_{6}{216}=3$

Find the value of $x$ if $\log \sin \sqrt x=0$

$\log \sin \sqrt x=0\\\sin \sqrt x=1\\\sqrt x=\dfrac \pi 2\\x=\dfrac{\pi ^2}4$

Evaluate: $\log_{4}\log_{3}\log_{2}{x}=0$

$\log_{4}\log_{3}\log_{2}{x}=0$

$\Rightarrow \log_{3}\log_{2}{x}={4}^{0}$

$\Rightarrow \log_{3}\log_{2}{x}=1$

$\Rightarrow \log_{2}{x}={3}^{1}$

$\Rightarrow \log_{2}{x}=3$

$\Rightarrow x={2}^{3}=8$

$\therefore\,x=8$

If $\log_{x}\dfrac 18=\dfrac{-3}{2}$ then x is equal to

$\log_x\dfrac{1}{8}=\dfrac{-3}{2}$

$\Rightarrow$

$x^{\log_x\tfrac{1}{8}}=x^{\tfrac{-3}{2}}$

$\Rightarrow$

$\dfrac{1}{8}=x^{\tfrac{-3}{2}}$

$\Rightarrow$

$x^{\tfrac{3}{2}}=8$

$\Rightarrow$

$\sqrt[3]{x^{3/2}}=\sqrt[3]{8}$

$\Rightarrow$

$x^{\tfrac{1}{2}}=2$

$\Rightarrow$

$\left(x^{\tfrac{1}{2}}\right)^2=(2)^2$

$\therefore$

$x=4$

If $\log_4x=12$ , then $\log_2 \dfrac x4$ is equal to

$\log_4x=12$ [ Given ]

$\Rightarrow$

$x=4^{12}$

$[\,\log_ax=b\Rightarrow x=a^b]$

Now,

$\Rightarrow$

$\log_2\left(\dfrac{x}{4}\right)=\log_2\left(\dfrac{4^{12}}{4}\right)$

$=\log_2\left(4^{11}\right)$

$=\log_2\left((2^2)^{11}\right)$ [ As

$4=2^2$ ]

$=\log_2\left(2^{22}\right)$ [ As

$(a^m)^n=a^{mn}$ ]

$=22$

$[\,\log_a a^b=b\,]$

$\therefore$

$\log_2\left(\dfrac{x}{4}\right)=22$

Simplify: ${a}^{\frac{\log_{b}{\left(\log_{b}{N}\right)}}{\log_{b}{a}}}$

${a}^{\frac{\log_{b}{\left(\log_{b}{N}\right)}}{\log_{b}{a}}}$

$={a}^{\log_{a}{\left(\log_{b}{N}\right)}}$ using

$\dfrac{\log_{a}{b}}{\log_{a}{c}}=\log_{c}{b}$

$=\log_{b}{N}$ using

${a}^{\log_{a}{N}}=N$

Solve: $\log_{e}\log_{5}\left[\sqrt{2x-2}+3\right]=0$

$\log_{e}\log_{5}\left[\sqrt{2x-2}+3\right]=0$

$\Rightarrow \log_{5}\left[\sqrt{2x-2}+3\right]={e}^{0}=1$

$\Rightarrow \log_{5}\left[\sqrt{2x-2}+3\right]=1$

$\Rightarrow \sqrt{2x-2}+3={5}^{1}=5$

$\Rightarrow \sqrt{2x-2}=5-3=2$

$\Rightarrow 2x-2=4$

$\Rightarrow 2x=4+2=6$

$\Rightarrow 2x=6$

$\Rightarrow x=\dfrac{6}{2}=3$

$\therefore\,x=3$

Find the value of $\log _8512+\log _28$

Let

$a=\log _8512+\log _28$

$=\log_88^3+\log_22^3$

$=3\log_88+3\log _22 \quad \dots ( n \log_ab=\log_ab^n)$

$a=3+3=6$

$\log _8512+\log _28=6$

Solve: $\log{\left(\log{x}\right)}+\log{\left(\log{{x}^{3}}-2\right)}=0$

$\log{\left(\log{x}\right)}+\log{\left(\log{{x}^{3}}-2\right)}=0$

$\Rightarrow \log{\left(\log{x}\right)\left(\log{{x}^{3}}-2\right)}=0$

$\Rightarrow {\left(\log{x}\right)\left(\log{{x}^{3}}-2\right)}={10}^{0}=1$

$\Rightarrow \log{x}\left(3\log{x}-2\right)=1$

$\Rightarrow 3{\left(\log{x}\right)}^{2}-2\log{x}-1=0$

$\Rightarrow 3{\left(\log{x}\right)}^{2}-3\log{x}+\log{x}-1=0$

$\Rightarrow 3\log{x}\left(\log{x}-1\right)+1\left(\log{x}-1\right)=0$

$\Rightarrow \left(\log{x}-1\right)\left(3\log{x}+1\right)=0$

$\Rightarrow \log{x}=1,\,\dfrac{-1}{3}$

$\therefore x={10}^{1},\,{10}^{\frac{-1}{3}}$

$\therefore\,x=10,\dfrac{1}{\sqrt[3]{10}}$

Solve for $x:\,\,\dfrac{\log_{10}{\left(x-3\right)}}{\log_{10}{\left({x}^{2}-21\right)}}=\dfrac{1}{2}$

$\dfrac{\log_{10}{\left(x-3\right)}}{\log_{10}{\left({x}^{2}-21\right)}}=\dfrac{1}{2}$

$\Rightarrow \log_{\left({x}^{2}-21\right)}{\left(x-3\right)}=\dfrac{1}{2}$

$\Rightarrow x-3=\sqrt{{x}^{2}-21}$

By squaring both sides, we get

$\Rightarrow {\left(x-3\right)}^{2}={x}^{2}-21$

$\Rightarrow {x}^{2}-6x+9={x}^{2}-21$

$\Rightarrow -6x+9=-21$

$\Rightarrow -6x=-21-9=-30$

$\Rightarrow x=\dfrac{-30}{-6}=5$

$\therefore\,x=5$

Find ${x}^{\log{x}+4}=32$ where base of logarithm is $2$

${x}^{\log_{2}{x}+4}={2}^{5}$

Applying

$\log_{2}$ to both sides,

$\Rightarrow \left(\log_{2}{x}+4\right)\log_{2}{x}=5\log_{2}{2}$

$\Rightarrow \left(\log_{2}{x}+4\right)\log_{2}{x}=5$

$\Rightarrow {\left(\log_{2}{x}\right)}^{2}+4\log_{2}{x}-5=0$

Let

$t=\log_{2}{x}$

$\Rightarrow {t}^{2}+4t-5=0$

$\Rightarrow {t}^{2}+5t-t-5=0$

$\Rightarrow t\left(t+5\right)-1\left(t+5\right)=0$

$\Rightarrow \left(t+5\right)\left(t-1\right)=0$

$\therefore\,t=1,\,-5$

$\Rightarrow\,\log_{2}{x}=1,\,-5$

$\Rightarrow x={2}^{1},\,{2}^{-5}$

$\therefore\,x=2,\,\dfrac{1}{32}$

NTR says that the degree of $(x^5-5)(x^2+3)$ isDo you agree with him?How?

No.

As,

$(x^5-5)(x^2+3)\\=x^{7}+3x^5-5x^2-15$

Max(power x)

$=17$

Simplify: $\displaystyle \frac { 2\sqrt { 30 } }{ \sqrt { 6 } } -\frac { 3\sqrt { 140 } }{ \sqrt { 28 } } +\frac { \sqrt { 275 } }{ \sqrt{55} }$

$\dfrac{2\sqrt{30}}{\sqrt{6}}$ -

$\dfrac{3\sqrt{140}}{\sqrt{28}}$ +

$\dfrac{275}{\sqrt{55}}$

$= \dfrac{2\sqrt{2\times 3\times 5}}{\sqrt{2\times 3}} - \dfrac{3\sqrt{2\times 2\times 5\times 7}}{\sqrt{2\times 2\times 7}} + \dfrac{\sqrt{5\times 5\times 11}}{\sqrt{5\times 11}}$

$= 2\sqrt{5} - 3\sqrt{5} + \sqrt{5}$

$= 3\sqrt{5} - 3\sqrt{5}$

$= 0$

log 5- log 7 =?

$\textbf{Step 1 : Use properties of logarithm}$

$\text{Given ,} \log 5 - \log 7$

$= \log (\dfrac{5}{7})$

$\boldsymbol {[\because \log m - \log n = \log (\dfrac{m}{n}) ]}$

$\textbf{Hence , } \boldsymbol { \log 5 - \log 7 } \textbf{ is equal to } \boldsymbol{ \log (\dfrac{5}{7}) }$

If $a^x=b, b^y=a$ then show that $xy=1$ .

$\textbf{Use laws of exponents and solve}$

$\text{The two given equations are}\ a^x=b\ \text{and}\ b^y=a$

$\text{Let us consider the second equation}\ b^y=a$

$\text{We can rewrite the equation as}\ b=a^{\frac{1}{y}}$

$\text{Use this value of b in}, a^x=b$

$\large{a^x=a^{\frac{1}{y}}}$

$\text{As bases are same, the powers can be equated to each other,}$

$x=\cfrac{1}{y}$

$xy=1$

$\textbf{Hence the given expression has been proved}$

In exponential form $729=3^a$ , what is the value of $a$ ?

We have,

$729$

$=9^3$

$=(3^2)^3$

$=3^6$ .
comparing it with

$3^a$ we get

$a=6.$

The exponential form of $625$ is $5^k$ then value of $k$ is ___.

Given,

$625$

We can write

$625$ as

$5 \times 5 \times 5 \times 5$

$625 = 5^4$

Write the given expression in exponential form.
$4a^3\times 6ab^2\times c^2$ .

$4a^{3} = 4\times a \times a \times a$

$6ab^{2} = 6\times a\times b \times b$ .

$c^{2} = c \times c$ .

Thus,

$4a^{3} \times 6a b^{2} \times c^{2} =$

$4\times a \times a \times a \times 6\times a\times b \times b \times c \times c$

$= 4\times 6 \times a \times a \times a \times a\times b \times b \times c \times c$

$= 24 \times a \times a \times a \times a\times b \times b \times c \times c$

$= 24\times a^{1+1+1+1}\times b^{1+1}\times c^{1+1}$

$= 24a^{4}b^{2}c^{2}$ .

Write the following in exponential form.
(i) $log_{10}100=2$
(ii) $log_{5}25=2$
(iii) $log_{2}2=1$

$log_{10}100=2$

in exponential form

$10^2=100$

ii)

$log_525=2$

in exponential form

$5^2=25$

iii)

$log_22=1$

in exponential form

$2^1=2$ .

Simplify: $\log_{\frac{1}{3}}{\sqrt[4]{729\sqrt[3]{{9}^{-1}{27}^{\frac{-4}{3}}}}}$

$\log_{\frac{1}{3}}{\sqrt[4]{729\sqrt[3]{{9}^{-1}{27}^{\frac{-4}{3}}}}}$

$=\log_{\frac{1}{3}}{\sqrt[4]{729\sqrt[3]{{3}^{-2}{3}^{\frac{-12}{3}}}}}$

$=\log_{\frac{1}{3}}{\sqrt[4]{729\sqrt[3]{{3}^{-2-\frac{12}{3}}}}}$

$=\log_{\frac{1}{3}}{\sqrt[4]{729\sqrt[3]{{3}^{-2-4}}}}$

$=\log_{\frac{1}{3}}{\sqrt[4]{729\sqrt[3]{{3}^{-6}}}}$

$=\log_{\frac{1}{3}}{\sqrt[4]{729{{3}^{-6}}^{\frac{1}{3}}}}$

$=\log_{\frac{1}{3}}{\sqrt[4]{729{{3}^{-2}}}}$

$=\log_{\frac{1}{3}}{\sqrt[4]{729\times\dfrac{1}{9}}}$

$=\log_{\frac{1}{3}}{\sqrt[4]{81}}$

$=\log_{\frac{1}{3}}{\sqrt[4]{{3}^{4}}}$

$=\log_{\frac{1}{3}}{{\left({3}^{4}\right)}^{\frac{1}{4}}}$

$=\log_{\frac{1}{3}}{3}$

$=\log_{{3}^{-1}}{3}$

$=-1$

The exponential form of $512$ is $2^k$ then value of $k$ is ___.

Given,

$512$

We can write

$512$ as

$2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$2^9$

Solve: ${9}^{1+\log{x}}-{3}^{1+\log{x}}-210=0$ where base of $\log$ is $3$

${9}^{1+\log{x}}-{3}^{1+\log{x}}-210=0$

$\Rightarrow {3}^{2\left(1+\log{x}\right)}-{3}^{1+\log{x}}-210=0$

Let

$t={3}^{1+\log{x}}$

$\Rightarrow {t}^{2}-t-210=0$

$\Rightarrow {t}^{2}-15t+14t-210=0$

$\Rightarrow t\left(t-15\right)+14\left(t-15\right)=0$

$\Rightarrow \left(t-15\right)\left(t+14\right)=0$

$\Rightarrow t=15,\,-14$

$\Rightarrow {3}^{1+\log{x}}=15,\,{3}^{1+\log{x}}=-14$

Taking

$\log$ both sides, we get

$\Rightarrow \log_{3}{\left({3}^{1+\log{x}}\right)}=\log_{3}{15},\,\,\log_{3}{\left({3}^{1+\log{x}}\right)}=\log_{3}{-14}$ is not defined.

$\therefore\,\,\log_{3}{\left({3}^{1+\log_{3}{x}}\right)}=\log_{3}{15}$

$\Rightarrow \,\,1+\log_{3}{x}=\log_{3}{\left(3\times 5\right)}$

$\Rightarrow \,\,1+\log_{3}{x}=\log_{3}{3}+\log_{3}{5}$

$\Rightarrow \,\,1+\log_{3}{x}=1+\log_{3}{5}$

$\Rightarrow \,\log_{3}{x}=\log_{3}{5}$

$\therefore\,x=5$

If $\log_{10}{\left({x}^{2}-12x+36\right)}=2$

$\log_{10}{\left({x}^{2}-12x+36\right)}=2$

$\Rightarrow \,{x}^{2}-12x+36={10}^{2}=100$

$\Rightarrow \,{x}^{2}-12x+36-100=0$

$\Rightarrow \,{x}^{2}-12x-64=0$

$\Rightarrow \,{x}^{2}-16x+4x-64=0$

$\Rightarrow \,x\left(x-16\right)+4\left(x-16\right)=0$

$\Rightarrow \,\left(x-16\right)\left(x+4\right)=0$

$\therefore\,x=-4,16$

Find the $5$ th root of $.003$ , having given $log 3=.4771213$ and $log 312936=5.4954243$ .

1662348_1731030_ans_a98cebfab10c4e6c9d732b8fa668b024.png

Given $log 11=1.0413927$ and $log 13=1.1139434$ , find the values of
$(1)$ $log 1.43$ , $(2)$ $log 133.1$ , $(3)$ $log \sqrt[4]{143}$ , and $(4)$ $log \sqrt[3]{.00169}$ .

1662359_1731025_ans_c7b788622c8546beb459da828b8b1ce0.png

Given $log 4=.60206$ and $log 3=.4771213$ , find the logarithms of $.8, .003, .0108$ , and $(.00018)^{\dfrac{1}{7}}$ .

1662378_1731023_ans_b314a49f3b2e450cbe5644bd63a23297.png

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log \sqrt[3]{12}.$

$\log2 =.3010300$ ,

$\log 3=.4771213$

$\log\sqrt[3]{12}=\dfrac{1}{3}(\log 2+\ log 2+\ log 3)$

$=\dfrac{1}{3}(.3010300+.3010300+.4771213)$

$=\dfrac{1}{3}(1.0791813)$

$=.3597271$

Find the value of $(1)$ $7^{\dfrac{1}{7}}$ , $(2)$ $(84)^{\dfrac{2}{5}}$ , and $(3)$ $(.021)^{\dfrac{1}{5}}$ , having given
$log 2=.30103$ , $log 3=.4771213$ ,
$log 7=.8450980, log 132057=5.1207283$ ,
$log 588453=5.7697117$ , and $log 461791=5.6644438$ .

1662342_1731035_ans_3827a72a437149af93ff598363442674.png

Find the numerical value of the logarithms of $7, 11$ and $13$ ; given $\mu=\cdot 43429448, \log 2=\cdot 30103000$ .

$\log 7 = 0.8451 \\ \log 11 = 1.0414 \\ \log 13 = 1.1139$

Solve
$\dfrac{16\times 10^2\times 64}{2^4\times 4^2}$

$\cfrac{16\times 10^2 \times 64}{2^4 \times 4^2}=\cfrac{4^2\times 10^2 \times 4^3}{2^4 \times 4^2}$

Using

$a^m \div a^n =(a)^{m-n}$ and

$a^m \times a^n =a^{m+n}$

$\cfrac{16\times 10^2 \times 64}{2^4 \times 4^2}=(4)^2\times 10^2\times 2^{-4}\times (4)^3 \times 4^{-2}$

$=(4)^3\times 10^2 \times 2^{-4}$

$=(2^2 )^3\times 10^2 \times 2^{-4}$

$=2^6 \times 10^2 \times 2^{-4}$

$=2^2 \times 10^2 =4\times 100 =400$

While studying her family's history. Shikha discovers records of ancestors she has had in the past $12$ generations. She started to make a diagram to help her figure this out. The diagram soon become very complex.
Make a graph showing the number of ancestors in each of the $12$ generations.

From question it is clear every past generation has twice as many people as this generation.

Let, Shikha's generation be past generation number

$0$ , So, her parent's will be past generation number

$1$ , her grandparent's will be

$2$ and so on.

So, this is the table showing number of ancestors in each generation

past generation number	Number of ancestors in that generation= $2\times$ number of ancestors in previous past generation
0	$1$
1	$12\times1=2$
2	$2\times2=2^2=4$
3	$2\times2^2=2^3=8$
4	$2\times2^3=2^4=16$
5	$2\times2^4=2^5=32$
6	$2\times2^5=2^6=64$
7	$2\times2^6=2^7=128$
8	$2\times2^7=2^8=256$
9	$2\times2^8=2^9=512$
10	$2\times2^9=2^{10}=1024$
11	$2\times2^{10}=2^{11}=2048$

Also, graph of past generation number and ancestors is shown from the table.

Two machines can be hooked together. When something is sent through this hook up, the output from the first machine becomes the input for the second.
When two machines hooked together do the same work as $(\times 10^2)$ machine does? Is there more than one arrangement of two machines that will work?

So, if

$x$ is input for

$\times10^2$ machine, then

$x\times 10^2=x100$ is output.

To get the same work as of

$(\times 10^2)$ machine, hook

$( \times 2^2)$ and

$( \times 5^2)$ machine.

So, for

$x$ inupt in

$1^{st}$ machine, we get,

$x\times2^2 \times 5^2 =x\times 4\times 25=100x$ output through second machine

Also, to get the same work as of

$(\times 10^2)$ machine, hook

$( \times 50^1)$ and

$( \times 2^1)$ machine.

So, for

$x$ inupt in

$1^{st}$ machine, we get,

$x\times50^1 \times 2^1=x\times50 \times 2=100x$ output through second machine.

From these two examples it is clear that more than one arrangements will work.

Arrangements possible will be factors of

$100$

For the following repeater machines, how many times the base is applied and how much the total stretch is ?

Repeater Machine is a hypothetical machine which automatically enlarges items several times.

So, from figure,

In the first machine that is

$(a)$ base is

$100$ because

$(\times100)$ is written and it is applied

$2$ times as it is written next to it.

So, output of the machine will be

$(\times 100^2 )=\times100\times100=\times10000$ stretch.

In the first machine that is

$(b)$ base is

$7$ because

$(\times7)$ is written and it is applied

$5$ times as it is written next to it.

So, output of the machine will be

$(\times 7^5 )=\times7\times7\times7\times7\times7=\times16807$ stretch.

In the first machine that is

$(c)$ base is

$5$ because

$(\times5)$ is written and it is applied

$7$ times as it is written next to it.

So, output of the machine will be

$(\times 5^7 )=\times5\times5\times5\times5\times5=\times3125$ stretch.

Find a single repeater machine that will do the same work as the hook -up.

A machine

$(\times x^y)$ with input

$a$ will give an output

$ax^y$

When, multiple such machines are hooked up their stretching powers are multiplied.

Machine

$1$ is

$(\times 2^2)$

Machine

$2$ is

$(\times 2^3)$

Machine

$3$ is

$(\times 2^4)$

As per the law of exponent-

$a^m \times a^n =a^{m+n}$

So,

The work that can be performed by the given repeater machine is equal to

$2^2 \times 2^3 \times 2^4 =2^9$ .

Hence,

$(\times 2^9)$ is the single machine that can perform the same amount of work as given hooked up machines

Find a single machine that will do the same job as the given hook-up.
$A(\times 2^3)$ machine followed by $(\times 2^{-2})$ machine.

If input to

$a$ is given to

$\times x^y$ machine then output is

$ax^y$

When, two such machines are hooked up their stretching power is multplied.

So, If

$( \times 2^3)$ machine is followed by

$(\times 2^{-2})$ machine,

the result will be

$2^3\times2^{-2}=2\times 2\times 2\times \dfrac 12 \times \cfrac12=2^1=2$

Thus, a single machine

$( \times 2^1)$ will also work as the given hook-up

Find a single machine that will do the same job as the given hook-up.
$A(\times 5^{99})$ machine followed by $(5^{-100})$ machine.

A machine

$(\times x^y)$ with input

$a$ will give an output

$ax^y$

When, two such machines are hooked up their stretching powers are multiplied.

$(\times 5^{99})$ machine is followed by

$( \times 5^{-100})$ machine-

Therefore, the result will be

$=\times5^{99}\times5^{-100}=\times5^{99}\times \dfrac{1}{5^{100}}=\times \dfrac 15=\times 5^{-1}$

So, a single machine

$\left(\times\cfrac15^1\right)$ or machine

$(\times 5^{-1})$ will do same job as that of given hooked up machines

Shikha has an order from a golf course designer to put palm trees through $a(\times 2^3)$ machine and then through $(\times 3^3)$ machine. She thinks she can do the job with a single repeater machine. What single repeater machine should she use?

A machine

$(\times x^y)$ with input

$a$ will give an output

$ax^y$

When, multiple such machines are hooked up their stretching powers are multiplied.

Machine

$1$ is

$(\times 2^3)$

Machine

$2$ is

$(\times 3^3)$

As per the law of exponent-

$a^m \times a^n =a^{m+n}\ \&\ a^mb^m=(ab)^m$

So,

The work that can be performed by the repeater machine is equal to

$2^3 \times 3^3=6^{3}$ .

So, she can use

$( \times 6^3)$ single machine for the same purpose.

Find a single repeater machine that will do the same work as the hook -up.

A machine

$(\times x^y)$ with input

$a$ will give an output

$ax^y$

When, multiple such machines are hooked up their stretching powers are multiplied.

Machine

$1$ is

$\left(\times\left(\cfrac12\right)^2\right)$

Machine

$2$ is

$\left(\times\left(\cfrac13\right)^2\right)$

As per the law of exponent-

$a^m \times a^n =a^{m+n}$

So,

The work that can be performed by the repeater machine is equal to

$\times\left( \dfrac 12 \right)^2\times \left( \dfrac 13 \right)^2= \times\left( \dfrac 12 \right)\times \left( \dfrac 12 \right)\times \left( \dfrac 13 \right)\times \left( \dfrac 13 \right)=\times\dfrac{1}{2\times 2\times 3\times 3}=\times\dfrac{1}{(6)^2}$ .

Hence,

$\left(\times\left(\cfrac16\right)^2\right)$ is a single machine the can perform the same amount of work as given hooked up machines.

For the hook-up, determine whether there is a single repeater machine that will do the same work. If so, describe or draw it.

A machine

$(\times x^y)$ with input

$a$ will give an output

$ax^y$

When, multiple such machines are hooked up their stretching powers are multiplied.

Machine

$1$ is

$(\times 7^3)$

Machine

$2$ is

$(\times 7^2)$

As per the law of exponent-

$a^m \times a^n =a^{m+n}$

So,

The work that can be performed by the repeater machine is equal to

$7^3 \times 7^2=7^{5}$ .

So, Yes,

$(\times 7^{5})$ is a single machine the can perform the same amount of work as given hooked up machines.

It is described in diagram with

$7$ as base and

$5$ as power.

1678802_1793378_ans_65a575fa165247fea39287e664badb6d.png

For the hook-up, determine whether there is a single repeater machine that will do the same work. If so, describe or draw it.

A machine

$(\times x^y)$ with input

$a$ will give an output

$ax^y$

When, multiple such machines are hooked up their stretching powers are multiplied.

Machine

$1$ is

$(\times 0.5^2)$

Machine

$2$ is

$(\times 0.5^3)$

As per the law of exponent-

$a^m \times a^n =a^{m+n}$

So,

The work that can be performed by the repeater machine is equal to

$0.5^2 \times 0.5^3=0.5^{5}$ .

It is described in a figure with

$0.5$ as base and

$5$ as power.

1678910_1793381_ans_26b30dc28f124206a76dd62f9e9266a6.png

Find a single repeater machine that will do the same work as the hook -up.

A machine

$(\times x^y)$ with input

$a$ will give an output

$ax^y$

When, multiple such machines are hooked up their stretching powers are multiplied.

Machine

$1$ is

$(\times 3^y)$

Machine

$2$ is

$(\times 3^y)$

As per the law of exponent-

$a^m \times a^n =a^{m+n}$

So,

The work that can be performed by the repeater machine is equal to

$3^y \times 3^y=3^{2y}$ .

Hence,

$(\times 3^{2y})$ is a single machine the can perform the same amount of work as given hooked up machines.

Find the value of:
$(-3)^{2} \times 5^2$

$(-3)^{2} \times 5^2= (-3) \times (-3) \times 5 \times 5 =9 \times 25 = 225$

The diameter of the Sun is $1.4\times 10^9\ m$ and the diameter of the Earth is $1.2756\times 10^7\ m$ . Compare their diameters by division.

Given,

Sun's diameter

$=1.4\times 10^9\ m$

Earth's diameter

$=1.2756\times 10^7\ m$

For comparing, we need to change the diameters of both sun and earth in the same powers of

$10$ , i.e

Diameter of earth=

$1.2756\times 10^7=0.012756\times 10^9\ m$

Now, if the diameter of sun is divided by the diameter of the earth=

$\dfrac{1.4\times 10^9}{0.012756\times 10^9}\approx110$

Therefore, the sun's diameter is approximately

$110$ times bigger than the earth's diameter.

Simplify and write in exponential form:
$\dfrac{9^8 \times (x^2)^5}{(27)^4 \times (x^3)^2}$

$\dfrac{9^8 \times (x^2)^5}{(27)^4 \times (x^3)^2}$

$=\dfrac{(3^2)^8 \times x^{2\times 5}}{(3^3)^4 \times x^{3 \times 2}}$

$=\dfrac{3^{16} \times x^{10}}{3^{12} \times x^6}$

$=3^{16-12} \times x^{10-6}$

$=3^4 \times x^4$

$=(3x)^4$

Simplify:
$\dfrac{7^3 \times 11^4 \times 13^0 }{7^2 \times 11^2}$

$\dfrac{7^3 \times 11^4 \times 13^0 }{7^2 \times 11^2}$

$=7^{3-2}\times 11^{4-2} \times 13^0$

$=7^1 \times 11^2 \times 1$

$=7 \times 11 \times 11$

$=847$

Find $x$ .
$\left( \dfrac 25\right)^{2x+6}\times \left( \dfrac 25\right)^{3}=\left( \dfrac 25\right)^{x+2}$

$\left( \cfrac 25\right)^{2x+6}\times \left( \cfrac 25\right)^{3}=\left( \cfrac 25\right)^{x+2}$

As per the law of exponents:

$a^m \times a^n =(a)^{m+n}$

Therefore,

$\left( \cfrac 25\right)^{2x+6+3}=\left( \cfrac 25 \right)^{x+2}$

When we compare the powers of

$\left( \dfrac 25\right)$ , we get:

$2x+6+3=x+2$

$\Rightarrow2x+9=x+2$

$\Rightarrow x=-7$

Simplify:
$\dfrac{[(-5)^3]^4 \times 8^2}{4^3 \times (25)^5}$

$\dfrac{[(-5)^3]^4 \times 8^2}{4^3 \times (25)^5}$

$=\dfrac{(-5)^{3\times 4}\times (2^3)^2 }{(2^2)^3 \times (5^2)^5}$

$=\dfrac{(-5)^{12} \times 2^6}{2^6 \times 5^{10}}$

$=\dfrac{(-1)^{12}(5)^{12}\times 2^6}{2^6\times 5^{10}}$

$=1\times 5^{12-10} \times 2^{6-6}$

$=5^2 \times 2^0$

$=25 \times 1$

$=25$

Simplify and write in exponential form:
$\dfrac{(-3)^5\times 8^3 \times 2^5}{3^2 \times 4^4}$

$\dfrac{(-3)^5\times 8^3 \times 2^5}{3^2 \times 4^4}$

$=\dfrac{(-1)^5 \times 3^5 \times (2^3)^3 \times 2^5}{3^2 \times (2^2)^4}$

$=\dfrac{-1 \times 3^5 \times 2^9 \times 2^5}{3^2 \times 2^8}$

$=-1[3^{5-2} \times 2^{9+5-8}]$

$=-1[3^3 \times 2^6]$

$=(-3)^3 \times 2^6$

$=(-27 \times 64)$

$=-1728$

Simplify:
$\dfrac{(-2)^3 \times (3x)^2 \times (-xy^3)}{3x^2y}$

$\dfrac{(-2)^3 \times (3x)^2 \times (-xy^3)}{3x^2y}$

$=\dfrac{(-2) \times (-2) \times (-2) \times (3)^{2} \times x^{2} \times (-x) \times y^3}{3 \times x^2 \times y}$

$=\dfrac{-8 \times 9x^2 \times (-1)x \times y^{3}}{3x^{2} \times y}$

$=\dfrac{+72}{3}xy^{3-1}$

$=24xy^2$

The left column of the chart lists the lengths of input chains of gold. Repeater machines are listed across the top. The other entries are the outputs you get when you send the input chain from that row the repeater machine from that column. Copy and complete the chart.
Input Length Repeater Machine
$x\ 2^3$
$40$ $125$
$2$ $162$

$81$

There on column on left which is input and

$3$ columns on right of that which represent output of each machine .

Stretching value of each machine is given in second row.

So, from third row we know that

$(\times 2^3=\times8)$ machine gives output

$40$

So, input of third row is

$\cfrac{40}{8}=5$

And, third machine gives output

$125$ ,

So stretching power of second machine is

$\cfrac{125}{5}=\times 25=\times5^2$

Now, from fourth row, output of second machine is

$162$ and input is

$2$

So, stretching power of third machine is

$\cfrac{162}2=\times81=\times3^4$

For , fifth row output of second machine

$(\times 3^4)$ is

$81$

So, input of fifth row is

$\cfrac{81}{3^4}=\cfrac{81}{81}=1$

We, have all inputs and all stretching power, so we can prepare table and fill remaining places:

So, table will be

Input length	machine $1$	machine $2$	machine $3$
	$(\times 2^3)$	$(\times3^4)$	$(\times 5^2)$
$5$	$40$	$405$	$125$
$2$	$16$	$162$	$50$
$1$	$8$	$81$	$25$

Simplify:
$\left( \dfrac{4}{13}\right)^{4}\times \left( \dfrac{13}{7}\right)^{2}\times \left( \dfrac{7}{4}\right)^{3}$

$\left( \dfrac{4}{13}\right)^{4}\times \left( \dfrac{13}{7}\right)^{2}\times \left( \dfrac{7}{4}\right)^{3}$

On simplifying equation, we get :

$\left( \dfrac{4}{13}\right)^{4}\times \left( \dfrac{13}{7}\right)^{2}\times \left( \dfrac{7}{4}\right)^{3}=\left( \dfrac {(4)^4}{(13)^4}\right)\times \left( \dfrac {(13)^2}{(7)^2}\right)\times \left( \dfrac {(7)^3}{(4)^3}\right)$

$=(4)^4\times (4)^{-3}\times (13)^2 \times (13)^{-4}\times (7)^3 \times (7)^{-2}$

Because

$a^{-m}=\dfrac{1}{a^m}$

$=(4)^{4-3}\times (13)^{2-4}\times (7)^{3-2}$

$a^m \times a^n =a^{m-n}$

$=(4)^1 \times (13)^{-2}\times (7)^1$

Again using,

$a^{-m}=\dfrac{1}{a^m}$

$=4\times \dfrac {1}{169}\times 7$

$=\cfrac{28}{169}$

Evalute
$2^3 \times 5^2$

$2^3 \times 4^2$
It can be written as

$= 2 \times 2 \times 2 \times 5 \times 5$
On further calculation

$=8 \times 25$

$= 200$

Evaluate
$2^3 \times 4^2$

$2^3 \times 4^2$
It can be written as

$= 2 \times 2 \times 2 \times 4 \times 4$
On further calculation

$=8 \times 16$

$= 128$

Simplify and write in exponential form:
$\dfrac{3^2 \times 7^8 \times 13^6}{21^2 \times 91^3}$

$\dfrac{3^2 \times 7^8 \times 13^6}{21^2 \times 91^3}$

$=\dfrac{3^2 \times 7^8 \times 13^6}{(3 \times 7)^2 \times (7\times 13)^3}$

$=\dfrac{3^2 \times 7^8 \times 13^6}{3^2 \times 7^2 \times 7^3 \times 13^3}$

$=3^{2-2}\times7^{8-2-3} \times 13^{6-3}$

$=3^0 \times 7^3 \times 13^3$

$=1 \times 7^3 \times 13^3$

$=(7\times 13)^3$

$=(91)^3$

Elaluate
$2^2 \times 3^3$

$2^2 \times 3^3$
It can be written as

$= 2 \times 2 \times 3 \times 3 \times 3$
On further calculation

$=4 \times 27$

$= 108$

Simplify and write in exponential form:
$(-3)^6 \times (-5)^6$

$(-3)^6 \times (-5)^6$

$=[(-3) \times (-5)]^6$

$=(15)^6$ (

$a^m \times b^m = (a\times b)^m$ )

$=15^6$

Simplify and write in exponential form:
$\bigg(\dfrac{3}{10}\bigg)^5 \times \bigg(\dfrac{2}{15}\bigg)^5$

$\bigg(\dfrac{3}{10}\bigg)^5$

$=\bigg(\dfrac{3}{10}\times \dfrac{2}{15}\bigg)^5$ (

$a^m \times b^m = (a\times b)^m$ )

$=\bigg(\dfrac{1}{25}\bigg)^5$

$=\bigg(\dfrac{1}{5 \times 5}\bigg)^5$

$=\bigg(\dfrac{1}{5^2}\bigg)^5$

$=\bigg(\dfrac{1}{5}\bigg)^{5 \times 2}$

$=\bigg(\dfrac{1}{5}\bigg)^{10}$

If $(9^n \times 3^5 \times 27^3)(3\times 81^4)=27$ , find $n$ .

Let us simplify the given expression,

$\dfrac{[(3^{2})^{n}\times 3^{5}\times (3^{3})^{3}]}{(3\times (3^{4})^{4})}=3^{3}$

$\dfrac{[3^{2n}\times 3^{5}\times 3^{9}]}{(3\times 3^{16})}=3^{3}$

$3^{2n+5+9-1-16}=3^{3}$

Now by comparing the powers, we get

$2n+5+9-1-16=3$

$2n-3=3$

$2n=3+3$

$2n=6$

$n=6/2$

$=3$

Evaluate
$3^3 \times 5^2$

1887836_1825482_ans_8b630da3ed0a4df5bebab754177c1ea8.jpg

Write the following in logarithmic form:
$\frac{1}{81}=3^{c}$ .

1764195_1835619_ans_84f106198a4f4b0eacdc61da1effa7ae.PNG

Write the following in logarithmic form:
$10=5^{b}$ .

Elaluate
$3^2 \times 4^2$

$3^2 \times 4^2$
It can be written as

$= 3 \times 3 \times 4 \times 4$
On further calculation

$=9 \times 16$

$= 144$

Evaluate
$(2 / 3)^3 \times (3/4)^2$

$(2 / 3)^3 \times (3/4)^2$
It can be written as

$= (2/3) \times (2/3) \times (2/3) \times (3/4) \times (3/4)$
On further calculations

$=\dfrac{ 8}{27} \times \dfrac{9}{16}$

$=\dfrac{1}{6}$

Elaluate
$(4 \times 3)^3$

$(4 \times 3)^3$
It can be written as

$= 4 \times 4 \times 4 \times 3 \times 3 \times 3$
On further calculation

$=64 \times 27$

$= 1728$

Evaluate
$(3 / 5)^2 \times (-2 / 3)^3$

$(3/ 5)^2 \times (-2 / 3)^3$
It can be written as

$= (3/5) \times (3/5) \times (-2/3) \times (-2/3) \times (-2/3)$
On further calculations

$=\dfrac{9}{25} \times\dfrac{-8}{27}$

$=\dfrac{-8}{75}$

Elaluate
$5^3 \times 2^4$

$5^3 \times 2^4$
It can be written as

$= 5 \times 5 \times 5 \times 2 \times 2 \times 2 \times 2$
On further calculation

$=125 \times 16$

$= 2000$

Write the following in logarithmic form:
$7=2^{x}$

Elaluate
$(5 \times 4)^2$

$(5 \times 4)^2$
It can be written as

$= 5 \times 5 \times 4 \times 4$
On further calculation

$=25 \times 16$

$= 400$

Evaluate
$(-3 / 4)^3 \times (2 / 3)^4$

$(-3/ 4)^3 \times (2 / 3)^4$
It can be written as

$= (-3/4) \times (-3/4) \times (-3/4) \times (2/3) \times (2/3) \times (2/3) \times(2/3)$
On further calculations

$= \dfrac{-27}{64} \times \dfrac{16}{81}$

$=\dfrac{-1}{12}$

Solve the following:
$\log _{10}100000=z$

1764226_1835647_ans_3bce40db0ba44462ab6578b73e86b1c6.PNG

Write the following in exponential form:
$\log _{10}100=2$

1764207_1835629_ans_0df8924922a948fba8e2dfa02b9882c0.PNG

We have $\log _{2}32$ . Show that we get the same result by writing $32=2^{5}$ and then using power rules. Verify the answer.

Write the following in logarithmic form:
$\dfrac{1}{257}=4^{a}$

1764201_1835623_ans_112f99d6a6c449c7a6e03e5a92b71236.PNG

Write the following in exponential form:
$\log _{2}2=1$

1764216_1835633_ans_986bc87112954dfab8e4b51c70ef6afe.PNG

Solve the following:
$\log _{2}16=2$ $\therefore x^{2}=16\Rightarrow x=\pm 4$
Is it correct or not?

$\log _{x}16=2$

$x^{2}=4^{2},\therefore x=\pm 4$
No,

$x=\pm 4$ is not correct, because

$x$ is base to logarithm,

$'x'$ value never be negative. Logarithm can be equal to

$-ve$ number but base of logarithm can't be

$-ve$ .

Solve the following:
$\log _{2}32=x$

1764220_1835643_ans_4ce8635ce8b442c3aad07275db391317.PNG

Solve the following:
$\log _{5}625=y$

1764223_1835645_ans_af36fb76f9c342f7a800fde0ce77760b.PNG

Write the following in logarithmic form:
$100=10^{z}$

1764198_1835621_ans_d9da1e61783644f28a0f98143a600aa1.PNG

Write the following in exponential form:
$\log _{5}25=2$

1764211_1835630_ans_7bed1692c51c4be8ac8a65ff789f0435.PNG

Determine the value of the following:
$\log _{81}3$

1764292_1835813_ans_3f24977143c84e79921721fce8449d74.PNG

Determine the value of the following:
$\log _{2}\left ( \frac{1}{16} \right )$ .

$\log _{2}\left ( \frac{1}{16} \right )$

$=\log _{2}\left ( \frac{1}{2} \right )^4$

$=\log _{2}\left ( 2 \right )^{-4}$

$=-4\ \log _{2}\left ( 2 \right )$

$=-4$ .

Determine the value of the following:
$\log _{2}512$ .

1762653_1835820_ans_856d89b36d7e4dca91788449ab4a26b8.PNG

Determine the value of the following:
$\log _{10}0.01$ .

Find the value of:
$\log _{2}32$

1764278_1835789_ans_ea3ffa6ac2b4472d9a81491093b5f055.PNG

Determine the value of the following:
$\log _{7}1$ .

1764314_1835815_ans_e51a16690eb44ae5960790b9f7ec44be.PNG

Find the value of:
$\log _{\dfrac{2}{3}}\dfrac{8}{27}$ .

1764282_1835794_ans_472a75cd83174c4e83b8e45bf7b5aa8b.PNG

Determine the value of the following:
$\log _{\frac{2}{3}}\left ( \frac{8}{27} \right )$ .

Find the value of:
$\log _{10}0.001$

1764281_1835792_ans_58c50d56e4194e69ad86b62fc1f57b28.PNG

Determine the value of the following:
$\log _{25}5$

1764289_1835810_ans_04814009e3294bbc9c1d22fd8f291010.PNG

Express each of the following in logarithmic form:
$10^{-3}=0.001$

$10^{-3}=0.001$

$\Rightarrow \log_{10}0.001=-3...................[a^{b}]=c\Rightarrow \log_ac=b]$

Express each of the following in logarithmic form:
$5^{3}=125$

$5^{3}=125$

$\Rightarrow \log_{5}125=3......................[a^{b}]=c\Rightarrow log_ac=b]$

Expand the following:
$\log \left ( \dfrac{128}{625} \right )$ .

1762717_1835844_ans_651cc6be076a4690aee949412c55c893.PNG

Evaluate the following in terms of $x$ and $y$ , if it is given that $x=\log _{2}3$ and $y=\log _{2}5$ .
$\log _{2}60$ .

1762702_1835838_ans_e438b2622481453d8e0d5f7e78be5b06.PNG

Evaluate the following in terms of $x$ and $y$ , if it is given that $x=\log _{2}3$ and $y=\log _{2}5$ .
$\log _{2}7.5$ .

1762698_1835836_ans_bca91a7f8ef64d819b35160f430aa0bb.PNG

Evaluate the following in terms of $x$ and $y$ , if it is given that $x=\log _{2}3$ and $y=\log _{2}5$ .
$\log _{2}6750$ .

1762706_1835839_ans_e25d6577e506483dbf9d6c17c384443c.PNG

Evaluate the following in terms of $x$ and $y$ , if it is given that $x=\log _{2}3$ and $y=\log _{2}5$ .
$\log _{2}15$ .

1762689_1835835_ans_b51329add54243a7abced6d993de95b4.PNG

Determine the value of the following:
$2^{2+\log _{2}3}$ .

Find the logarithm of :
100 to the base 10

Let

$\log_{10}100=x$

$\therefore 10^{x}=100$

$\Rightarrow 10^{x}=10\times 10$

$\Rightarrow 10^{x}=10^{2}$

$\Rightarrow X =2[If \ a^{m}=a^{n};then \ m = n]$

$\therefore \log_{10}100=2$

Express the each of following in exponential form:
$\log _{a}A = x$

$Log_{a} A = x$

$\Rightarrow a^{x}=A \ \ \ [log_{a}c=b\Rightarrow a^{b}=c]$

Express the each of following in exponential form:
$log_{10}0.01=-2$

$log_{10}0.01=-2$

$\Rightarrow 10^{-2}=0.01[log_{a}c=b\Rightarrow a^{b}=c]$

Find the logarithm of :
0.01 to the base 10

$\log_{10}0.01=x$

$\therefore 10^{x}=0.01$

$\Rightarrow 10^{x}=\dfrac{1}{10}$

$\Rightarrow 10^{x}=10^{-1}$

$\Rightarrow =-1[Ifa^{m}=a^{n};then \ m = n]$

$\therefore log_{10}0.1=-1$

Find the logarithm of :
32 to the base 4

$\log_{4}32=x$

$\therefore 4^{x}=32$

$\Rightarrow (2^{2})^{x}=2\times 2\times 2\times 2\times 2$

$\Rightarrow 2^{2x}=2^{5}$

$\Rightarrow 2x=5.......[If \ a^{m}=a^{n};then \ m = n]$

$\Rightarrow x=\dfrac{5}{2}$

$\therefore \log_{4}32=\dfrac{5}{2}$

Solve for x :
$\log_{10} x = -2$

$\log_{10}x=-2$

$\Rightarrow 10^{-2}=x[log,c=b\Rightarrow a^{b}=c]$

$\Rightarrow x=10^{-2}$

$\Rightarrow x = \dfrac{1}{10^{2}}$

$\Rightarrow x=\dfrac{1}{100}$

$\Rightarrow x = 0.01$

Express the each of following in exponential form:
$\log _{10}1 = 0$

$\log _{10}1 = 0$

$\Rightarrow 10^{0}=1[log_{a}c=b\Rightarrow a^{b}=c]$

Express each of the following in logarithmic form:
$3^{-2}=\dfrac{1}{9}$

$3^{-2}=\dfrac{1}{9}$

$\Rightarrow \log_{3}\dfrac{1}{9}=-2............................[a^{b}=c\Rightarrow \log_ac=b]$

Express the each of following in exponential form:
$\log_{8}0.125=-1$

$\log_{8}0.125=-1$

$\Rightarrow 8^{-1}=0.125.........................[\log_{a}c=b\Rightarrow a^{b}=c]$

Find the logarithm of :
0.001 to the base 10

$\log_{10}0.001=x$

$\therefore 10^{x}=0.001$

$\Rightarrow 10^{x}=\dfrac{1}{1000}$

$\Rightarrow 10^{x}=\frac{1}{10^{3}}$

$\Rightarrow 10^{x}=10^{-3}$

$\Rightarrow X=-3[Ifa^{m}=a^{n};thenm = n]$

$\therefore log_{10}0.001=-3$

Evaluate :
$log_{2}\dfrac{1}{8}$

Let

$\log_{2}\dfrac{1}{8}= x$

$\Rightarrow 2^{x}=\dfrac{1}{8}$

$\Rightarrow 2^{x}=\dfrac{1}{2\times 2\times 2}$

$\Rightarrow 2^{x}=\dfrac{1}{2^{3}}$

$\Rightarrow 2^{x}=2^{-3}$

$\Rightarrow x=-3$

Thus ,

$\log_{2}\dfrac{1}{8}=3$

Find x if :
$\log_{9}243=x$

Consider the equation

$\log_{9}243=x$

$\Rightarrow 9^{x}=243$

$\Rightarrow (3^{2})^{x}=3^{5}$

$\Rightarrow 3^{2x}=3^{5}$

$\Rightarrow 2x=5$

$\Rightarrow x=\dfrac{5}{2}$ and

$x=2\dfrac{1}{2}$

Find the logarithm
27 to the base 4

$\log_{9}27=x$

$\therefore 9^{x}=27$

$\Rightarrow (3\times 3)^{x}=3\times 3\times3$

$\Rightarrow (3^{2})^{x}=(3^{3})$

$\Rightarrow 3^{2x}=(3^{3})$

$\Rightarrow 2x=3 \ [If \ a^{m}=a^{n}; \ then \ m = n]$

$\Rightarrow x=\dfrac 32$

$\therefore \log_{4}27=\dfrac{3}{2}$

Evaluate :
$\log_{10}0.01$

$\log_{10}0.01=x$

$\Rightarrow 10^{x}=0.01$

$\Rightarrow 10^{x}=\dfrac{1}{100}$

$\Rightarrow 10^{x}=\dfrac{1}{10\times 10}$

$\Rightarrow 10^{x}=10^{-2}$

$\Rightarrow x=-2$

Thus ,

$\log_{10}0.01=2$

Find x if :
$\log_{4}32=x-4$

Consider the equation

$\log_{4}32=x-4$

$\Rightarrow 4^{x-4}=32$

$\Rightarrow (2^{2})^{x-4}=2^{5}$

$\Rightarrow 2^{(2x-8)}=2^{5}$

$\Rightarrow 2x-8=5$

$\Rightarrow 2x = 5+8$

$\Rightarrow 2x=13$

$\Rightarrow x=\dfrac{13}{2}$

$\Rightarrow x=6\dfrac{1}{2}$

Find the logarithm of :
0.125 to the base 2

$\log_{2}0.125=x$

$\therefore 2^{x}=0.125$

$\Rightarrow 2^{x}=0.125$

$\Rightarrow 2^{x}=\dfrac{125}{1000}$

$\Rightarrow 2^{x} =\dfrac{1}{8}$

$2^{x}=(2\times2\times2)^{-1}$

$2^{x}=(2^{3})^{-1}$

$2^{x}=2^{-3}$

$\Rightarrow x=-3[If \ a^{m}=a^{n};then \ m = n]$

$\therefore \log_{2}0.125=-3$

Find the logarithm
$\dfrac{1}{81}$ to the base 27

$\log_{27}\dfrac{1}{81}=x$

$\therefore 27^{x}=\dfrac{1}{81}$

$\Rightarrow (3\times 3\times 3)^{x}=\dfrac{1}{3\times 3\times 3\times 3}$

$\Rightarrow (3^{3})^{x}=\dfrac{1}{3^{4}}$

$\Rightarrow 3^{3x}=(3^{-4})$

$\Rightarrow 3x=-4 \ [If \ a^{m}=a^{n};then \ m = n]$

$\Rightarrow x=\dfrac{-4}{3}$

$\therefore \log_{27}\dfrac{1}{81}=\dfrac{-4}{3}$

Find the logarithm
$\dfrac{1}{16}$ to the base 4

$\log_{4}\dfrac{1}{16}=x$

$\therefore 4^{x}=\dfrac{1}{16}$

$\Rightarrow 4^{x}=\dfrac{1}{4\times 4}$

$\Rightarrow 4^{x}=(4\times 4)^{-1}$

$\Rightarrow 4^{x}=(4^{2})^{-1}$

$\Rightarrow 4^{x}=4^{-2}$

$\Rightarrow x=-2 \ [If \ a^{m}=a^{n};then \ m = n]$

$\therefore \log_{4}\dfrac{1}{16}=-2$

Evaluate :
$log_{5}1$

Let

$\log_{5}1=x$

$\Rightarrow 5^{x}=1$

$\Rightarrow 5^{x}=5^{0}$

$\Rightarrow x=0$

Thus ,

$\log_{5}1=0$

State , true or false :
$\log_{2}8=3$ and $\log _{8}2=\dfrac{1}{3}$

Consider the equation

$\log_{2}8=3$

$\Rightarrow 2^{3}=8....(1)$

Now consider the equation

$\log_{8}2=\dfrac{1}{3}$

$\Rightarrow 8^{\dfrac{1}{3}}=2$

$(2^{3})^{\dfrac{1}{3}}=2...(2)$

Both the equations ( 1) and (2) are correct

Thus the given statement

$\log_{2}8=3$ and

$\log _{8}2=\dfrac{1}{3}$ are true

If $\log_{10}2=a$ and $\log_{10}3=b;$ Express each of the following in terms of 'a' and 'b'
$\log 2.25$

$\log2.25=\log\dfrac{225}{100}$

$=\log\dfrac{25\times 9}{25\times 4}$

$=\log\dfrac{9}{4}$

$=\log\left ( \dfrac{3}{2} \right )^{2}$

$=2\log\left ( \dfrac{3}{2} \right )[n \log_{a}m=\log_{a}m^{n}]$

$=2(\log3-\log2)[\log_{a}m-\log_{a}n=\log _a\dfrac{m}{n}]$

$=2(b-a)[\because \log_{10}2=a \ and \ \log_{10}3=b]$

Prove that :
$2 \log\dfrac{15}{18}-\log\dfrac{25}{162}+\log\dfrac{4}{9}=\log2$

We need to prove that

$2 \log\dfrac{15}{18}-\log\dfrac{25}{162}+\log\dfrac{4}{9}=\log2$

$L.H.S= 2\log\dfrac{15}{18}-\log\dfrac{25}{162}+\log\dfrac{4}{9}$

$=\log \left ( \dfrac{15}{18} \right )^{2}-\log\dfrac{25}{162}+\log\dfrac{4}{9}$

$[n\log_{a}m=\log_{a}m^{n}]$

$=[\log\left ( \dfrac{15}{18} \right )\times \left ( \dfrac{15}{18} \right )]-\log\dfrac{25}{162}+\log\dfrac{4}{9}$

$=\log\left ( \dfrac{15}{18} \right )\times \left ( \dfrac{15}{18} \right )\times \dfrac{4}{9}-\log\dfrac{25}{162} [log_{a}m+log_{a}n=log_{a}mn]$

$=\log\dfrac{72}{36}$

$= \log 2$

$=R.H.S$

Evaluate :
$\log_{16}8=x$

$\log_{16}8=x$

$\Rightarrow 16^{x}=8$

$\Rightarrow (2\times 2\times 2\times 2)^{x}=2\times 2\times 2$

$\Rightarrow (2^{4})^{x}=2^{3}$

$\Rightarrow 4x=3$

$\Rightarrow x=\dfrac{3}{4}$

Thus,

$\log_{16}8=\dfrac{3}{4}$

State , true or false
$\dfrac{\log25}{\log5}=\log x$

Given that

$\dfrac{\log25}{\log5}=\log x$

$\Rightarrow \dfrac{\log5\times 5}{\log5}=\log x$

$\Rightarrow \dfrac{\log 5^{2}}{\log 5}=\log x$

$\Rightarrow \dfrac{2 \log5}{\log5}=\log x[\log_{a}m^{n}=nlog_{a}m]$

$\Rightarrow 2=\log_{10}x$

$\Rightarrow x=100$

Evaluate :
$\log_{5}125=x$

$\log_{5}125=x$

$\Rightarrow 5^{x}=125$

$\Rightarrow 5^{x}=5\times 5\times 5$

$\Rightarrow 5^{x}=5^{3}$

$\Rightarrow x=3$

Thus,

$\log_{5}125=3$

Solve for x :
$\dfrac{\log128}{\log32}=x$

$\dfrac{\log128}{\log32}=x$

$\Rightarrow x=\dfrac{\log128}{\log32}$

$\Rightarrow x=\dfrac{\log2\times 2\times 2\times 2\times 2\times 2\times 2}{\log2\times 2\times 2\times 2\times 2}$

$\Rightarrow x=\dfrac{\log2^{7}}{\log2^{5}}$

$\Rightarrow x=\dfrac{7\log2}{5\log2}[n\log_{a}m=\log_{a}m^{n}]$

$\Rightarrow x=\dfrac{7}{5}$

$\Rightarrow x=1.4$

If $\log_{10}2=a$ and $\log_{10}3=b;$ Express each of the following in terms of 'a' and 'b'
$\log2\dfrac{1}{4}$

$\log2\dfrac{1}{4}=\log\dfrac{9}{4}$

$=\log\left ( \dfrac{3}{2} \right )^{2}$

$=\log 3^2-\log 2^2$

$=2\log 3-2 \log 2$

$=2(b-a)[\because log_{10}2=a \ and \ log_{10}3=b]$

$=2b-2a$

If $\log_{10}2=a$ and $\log_{10}3=b;$ Express each of the following in terms of 'a' and 'b'
$\log 60$

$\log60=\log_{10}10\times2 \times 3$

$=\log_{10}10+\log_{10}2+\log_{10}3[\log_{a}mn=\log_{a}n+\log_{a}m]$

$=1+\log_{10}2+\log_{10}3[\because \log_{10}10=1]$

$=1+a+b[\because \log_{10}2=a \ and \ \log_{10}3=b]$

Solve for x :
$\dfrac{\log81}{\log27}=x$

$\dfrac{\log81}{\log27}=x$

$\Rightarrow x=\dfrac{\log81}{\log27}$

$\Rightarrow x=\dfrac{\log3\times 3\times 3\times 3}{\log3\times 3\times 3}$

$\Rightarrow x=\dfrac{\log3^{4}}{\log3^{3}}$

$\Rightarrow x=\dfrac{4\log3}{3\log3} [n\log_{a}m=\log_{a}m^{n}]$

$\Rightarrow x=\dfrac{4}{3}$

$\Rightarrow x=1\dfrac{1}{3}$

Evaluate :
$log_{0.5}16=X$

$log_{0.5}16=x$

$\Rightarrow 0.5^{x}=16$

$\Rightarrow \left ( \dfrac{5}{10} \right )^{x}=2\times 2\times 2\times 2$

$\Rightarrow \left ( \dfrac{1}{2} \right )^{x}=2^{4}$

$2^{-x}=2^{4}$

$\Rightarrow -x=4$

$\Rightarrow x = -4$

Thus ,

$\log_{0.5}16=-4$

If $\log 2 = 0.3010$ and $\log 3=0.4771;$ find the value of :
$\log 25$

$\log25=\log\left ( \dfrac{25\times 4}{4} \right )$

$=\log\left ( \dfrac{100}{4} \right )$

$.....[\log_{a} \dfrac mn=\log_{a}m-\log_{a}n]$

$=2\log10-2\log 2$

$=2-2(0.3010)[\because \log2=0.3010]$

$=1.398$

If $x = \log 0.6; y = \log 1.25$ and $z = \log 3-2 \log 2$ , find the values of :
$5^{x+y-z}$

Given that

$z=\log 3 -2 \log 2$

$=\log 3 - \log 2^{2}$

$=\log3-\log4$

$= \log \dfrac{3}{4}$

$=\log 0.75...(1)$

$x+y-z$

$=\log 0.6+\log 1.25-\log 0.75$

$=\log \dfrac{0.6\times 1.25}{0.75}$

$=\log\dfrac{0.75}{0.75}$

$=1$

$5^{x+y-z}=5^{0}=1$

If log 2 = 0.3010 and 0.4771; find the value of :
log 1.2

$\log1.2=\log\dfrac{12}{10}$

$=\log12-\log10.......[\log_{a}\dfrac{m}{n}=\log_{a}m-\log_{a}n]$

$=\log2\times 2\times 3-1[\because \log10=1]$

$=2 \log2+\log3-1$

$=2(0.3010)+0.4771-1\left [ {\because \log2=0.3010} \ {and \ \log3=0.4771} \right ]$

$=1.0791-1$

$=0.0791$

If $\log_{10}8=0.90;$ find the value of :
$\log 0.125$

Given that

$\log_{10}8=0.90;$

$\Rightarrow \log_{10}2\times 2\times 2=0.90$

$\Rightarrow \log_{10}2^{3}=0.90$

$\Rightarrow 3\log_{10}2=0.90$

$\Rightarrow \log_{10}2=\dfrac{0.90}{3}$

$\Rightarrow \log_{10}2=0.30...(1)$

$\log0.125=\log_{10}\dfrac{125}{1000}$

$=\log_{10}\dfrac{1}{8}$

$=\log_{10}\dfrac{1}{2\times 2\times 2}$

$=\log_{10}\left ( \dfrac{1}{2^{3}} \right )$

$=\log_{10}2^{-3}$

$=-3\times (0.30)[from(1)]$

$=-0.9$

If $\log_{10}8=0.90;$ find the value of :
$\log\sqrt{32}$

Given that

$\log_{10}8=0.90;$

$\Rightarrow \log_{10}2\times 2\times 2=0.90$

$\Rightarrow \log_{10}2^{3}=0.90$

$\Rightarrow 3 \log_{10}2=0.90$

$\Rightarrow \log_{10}2=\dfrac{0.90}{3}$

$\Rightarrow \log_{10}2=0.30...(1)$

$\log\sqrt{32}=\log_{10}(32)^{\tfrac{1}{2}}$

$\Rightarrow =\dfrac{1}{2}\log_{10}(32)$

$\Rightarrow =\dfrac{1}{2}\log_{10}(2\times 2\times 2\times 2\times 2)$

$\Rightarrow =\dfrac{1}{2}\log_{10}(2^{5})$

$\Rightarrow =\dfrac{1}{2}\times 5\log_{10}2$

$\Rightarrow =\dfrac{1}{2}\times 5(0.30)[]from (1)]$

$\Rightarrow 5\times 0.15$

$=0.75$

If $\log 2 = 0.3010$ and $\log 3=0.4771;$ find the value of :
$\log 3.6$

$\log3.6=\log\dfrac{36}{10}$

$=\log36-\log10.......[\log_{a}\dfrac{m}{n}=\log_{a}m-\log_{a}n]$

$=\log2\times 2\times 3 \times 3-1.......[\because \log10=1]$

$=2 \log2+2 \log3-1$

$=2(0.3010)+2 (0.4771)-1\left [ {\because \log2=0.3010} \ {and \ \log3=0.4771} \right ]$

$=1.5562-1$

$=0.5562$

If $\log 2 = 0.3010$ and $\log 3=0.4771;$ find the value of :
$\log 15$

$\log15=\log\left ( \dfrac{15 \times 10}{ 10} \right )$

$=\log\left ( \dfrac{15}{10} \right )+\log10$

$=\log3-\log2+1........[\because \log m-\log n=\log\left ( \dfrac{m}{n} \right )]$

$=0.4771-0.3010+1$

$=1.1761$

Given : $2 \log_{10}x+1=\log_{10}250,$
Find:
$\log_{10}2x$

$2 \log_{10}x+1=\log_{10}250,$

$\Rightarrow \log_{10}x^2+\log_{10}10=\log_{10}250$

$\Rightarrow \log_{10}x^2 \times 10=\log_{10}250$

$\Rightarrow x^2=25$

$\Rightarrow x=5$ as log is not defined for negative numbers

$\Rightarrow \log_{10}2x=\log_{10}2(5)=\log _{10}10$

$=1[\because log_{10}10=1]$

Find x , if :
$\log_{x}625=-4$

$\log_{x}625=-4$

$\Rightarrow 625=x^{-4}$ [Removing Logarithm]

$\Rightarrow 5^{4}=\left ( \dfrac{1}{x} \right )^{4}$

$\Rightarrow 5=\dfrac{1}{x}$ [Powers are same , bases are equal ]

$\Rightarrow x=\dfrac{1}{5}$

If $x = \log 0.6; y = \log 1.25$ and $z = \log 3-2 \log 2$ , find the values of :
$x+y-z$

Given that

$z=\log 3 - 2 \log2$

$=\log 3 - \log 2^{2}$

$=\log 3-\log4$

$=\log\dfrac{3}{4}$

$=\log 0.75...(1)$

$x+y-z$

$=\log 0.6+\log 1.25-\log 0.75$

$=\log \dfrac{0.6\times 1.25}{0.75}$

$=\log\dfrac{0.75}{0.75}$

$=1$

$=\log1$

$0...(2)$

Write the following in logarithm form:
$10^{4} = 10000$

Logarithm form of

$10^{4} = 10000$ is

$\log_{10} 10000 = 4$

Write the following in logarithm form:
$2^{10} = 1024$

Logarithm form of

$2^{10} = 1024$ is

$\log_{2} 1024 = 10$

If $\log_{\sqrt{27}}x=2\dfrac{2}{3}$ , find $x.$

$\log_{\sqrt{27}}x=2\dfrac{2}{3}$

$\therefore \log_{\sqrt{27}}x=\dfrac{8}{3}$

$\therefore x=(\sqrt{27})^{\tfrac{8}{3}}$

$\because \log_{a}x=b\Rightarrow x=a^{b}$

$\therefore x=\left ( 27^{\tfrac{1}{2}} \right )^{\tfrac{8}{3}}$

$\therefore x=\left ( 3^{\tfrac{3}{2}} \right )^{\tfrac{8}{3}}$

$\therefore x=3^{\dfrac{3}{2}\times \dfrac{8}{3}}$

$\therefore x=3^{4}$

$\therefore x=81$

Write the following in the power form :
$\log_{5} 25 = 2$

Exponential (power) form of

$\log_{5} 25 = 2$ is

$5^{2} = 25$

Solve for x
$\log_{x}15\sqrt{5}=2-\log_{x}3\sqrt{5}$

$\log_{x}15\sqrt{5}=2-\log_{x}3\sqrt{5}$

$\Rightarrow \log_{x}15\sqrt{5}+\log_{x}3\sqrt{5}=2$

$\Rightarrow \log_{x}(15\sqrt{5}\times 3\sqrt{5})=2$

$\Rightarrow \log_{x}225=2$

$\Rightarrow \log_{x}15^{2}=2$

$\Rightarrow 2 \log_{x}15=2$

$\Rightarrow \log_{x}15=1$

$\Rightarrow 15=x^1$

$\Rightarrow x=15$

Write the following in logarithm form:
$4^{3/2} = 8$

Logarithm form of

$4^{3/2} = 8$ is

$\log_{4} 8 = 3/2$ .

Solve for x , if
$\log_{x}49-\log_{x}7+\log_{x}\dfrac{1}{343}=2$

$\log_{x}49-\log_{x}7+\log_{x}\dfrac{1}{343}=2$

$\Rightarrow \log_{x}\dfrac{49}{7\times 343}=-2$

$\Rightarrow \log_{x}\dfrac{1}{49}=-2$

$\Rightarrow -\log_{x}49=-2$

$\Rightarrow 49=x^{2}$ [Removing logarithm]

$\therefore x=7$

Evaluate :
$\dfrac{\log_{5}8}{\log_{25}16\times \log_{10}10}$

$\dfrac{\log_{5}8}{\log_{25}16\times \log_{10}10}$

$=\dfrac{\dfrac{\log_{10}8}{\log_{10}5}}{\dfrac{\log_{10}16}{\log_{10}25}\times \dfrac{\log_{10}10}{\log_{10}100}}$

$=\dfrac{\dfrac{\log_{10}2^{3}}{\log_{10}5}}{\dfrac{\log_{10}2^{4}}{\log_{10}5^{2}}\times \dfrac{\log_{10}10}{\log_{10}10^{2}}}$

$=\dfrac{3\log_{10}2}{\log_{10}5}\times \dfrac{2\log_{10}5}{4\log_{10}2}\times \dfrac{2\log_{10}10}{\log_{10}10}$

$=3$

Write the following in logarithm form:
$10^{-3} = 0.001$

Logarithm form of

$10^{-3} = 0.001$ is

$\log_{10} 0.001 = – 3$

Write the following in logarithm form:
$5^{-2} =\frac{1}{25}$

Logarithm form of

$5^{-2} =\dfrac{1}{25}$ is

$\log_{5} (\dfrac{1}{25}) = -2$ .

If $\log_{125} P =\frac{1}{6}$ then find the value of $P$ .

$\log_{125} P =\frac{1}{6}$

$\Rightarrow P = (125)^{1/6}$

$\Rightarrow P = [(5)^{3}]^{1/6}$

$\Rightarrow P = 5^{1/2} = \sqrt{5}$
Hence, the value of

$P$ is

$\sqrt{5}$ .

Write the following in the power form :
$\log_{10} 0.1 = 1$

Exponential form of

$\log_{10} 0.1 = – 1$ is

$10^{-1} = 0.1$

Prove that:
$\log 630 = \log 2 + 2 \log 3 + \log 5 + \log 7$ .

$L.H.S. = \log 630$

$= \log (2 \times 3 \times 3 \times 5 \times 7)$

$= \log (2 \times 32 \times 5 \times 7)$

$= \log 2 + \log 3^{2} + \log 5 + \log 7$

$= \log 2 + 2 \log 3 + \log 5 + \log 7$

Hence Proved.

Write the following in the power form :
$log_{3} (\frac{1}{27} ) = -3$

Exponential form of

$\log_{3} (\dfrac{1}{27} ) = -3$ is

$3^{-3} =\dfrac{1}{27}$

Write the following in the power form :
$\log_{\sqrt{2}} 4 = 4$

Exponential (power) form of

$\log_{\sqrt{2}}4=4$ is

$(\sqrt{2})^{4}=4$

If $\log_{81} x =\frac{3}{2}$ , then find the value of $x$ .

$\log_{81} x =\dfrac{3}{2}$

$\Rightarrow x = (81)^{3/2}$

$= (9^{2})^{3/2} = (9)^{3} = 729$
Hence, the value of

$x$ is

$729$

If $\log_{4} m = 1.5$ , then find the value of $m$ .

$\log_{4} m = 1.5$

$\Rightarrow m = (4)^{1.5}$

$\Rightarrow m = (2^{2})^{1.5}$

$\Rightarrow m = 2^{3} = 8$
Hence, the value of

$m$ is

$8$ .

Prove that :
$\log_{4} [\log_{2}(\log_{2} (\log_{3} 81))] = 0$

We know that

$\log_{m} m^{n} = n \log_{m} m$
and

$\log_{m} m = 1$

$\therefore \log_{m} (m)_{n} = n * log_{m} m = n * 1 = n$

$\Rightarrow \log_{m} (m)^{n} = n$ …(i)

$L.H.S. = \log_{4} [\log_{2} (\log_{2} (\log_{3} 81))]$

$= \log_{4} [\log_{2} (\log_{2} (\log_{3} 3^{4} ))] (\because 81 = 3^{4} )$

$= \log_{4} [\log_{2} ((\log_{2}^{4}))$ [According to equal (i)]

$= \log_{4} (\log_{2} (\log_{2} 2^{2})) (\because 4 = 2^{2} )$

$= \log_{4} (\log_{2} 2)$ [According to equal (i)]

$= \log_{4} (1) (\because \log_{m} m = 1)$

$= 0 = R.H.S.$

Hence Proved.

Write the following in the power form :
$\log_{10} 0.001 = 3$

Exponential (power) form of

$\log_{10} 0.001 = – 3$ is

$10^{-3} = 0.001$

Write the following in the power form :
$\log_{3} 729 = 6$

Exponential (power) form of

$\log_{3} 729 = 6$ is

$3^{6} = 729$ .

Prove that:
$\log\frac{9}{14}+\log\frac{35}{24}-\log\frac{15}{16}=0$

$L.H.S.=\log\dfrac{9}{14}+\log\dfrac{35}{24}-\log\dfrac{15}{16}$

$=\log\dfrac{9}{14}+\log\dfrac{35}{24}+\log\dfrac{16}{15}$

$\left ( \because -\log\dfrac{a}{b}=\log\dfrac{b}{a} \right )$

$=\log\left ( \dfrac{9}{14}\times\dfrac{35}{24}\times\dfrac{16}{15} \right )$

$=\log\left ( \dfrac{3\times3\times5\times7\times8\times2}{2\times7\times8\times3\times3\times5} \right )$

$\log\left ( \dfrac{2\times3\times3\times5\times7\times8}{2\times3\times3\times5\times7\times8} \right )$

$\log1=0$ Hence proved.

If $\log 2 = 0.3010, \log 3 = 0.4771, \log 7 = 0.8451$ and $\log 11 = 1.0414$ , then find the value of the following :
$\log\left ( \frac{11}{7} \right )^{5}$

$\log\left ( \frac{11}{7} \right )^{5}=5\left ( \log\frac{11}{7} \right )$

$= 5(\log 11 – \log 7)$

$= 5(1.0414 – 0.8451)$

$= 5(0.1963)$

$= 0.9815$

If $\log 2 = 0.3010, \log 3 = 0.4771, \log 7 = 0.8451$ and $\log 11 = 1.0414$ , then find the value of the following :
$\log70$

$\log 70 = \log (7 \times 10)$

$= \log 7 + \log 10$

$= 0.8451 + 1$

$= 1.8451$

If $\log 2 = 0.3010, \log 3 = 0.4771, \log 7 = 0.8451$ and $\log 11 = 1.0414$ , then find the value of the following :
$\log36$

$\log 36 = \log (2 \times 2 \times 3 \times 3)$

$= \log (2^{2} \times 3^{2})$

$= \log 2^{2} + \log 3^{2}$

$= 2 \log 2 + 2 \log 3$

$= 2(\log 2 + \log 3)$

$= 2(0.3010 + 0.4771)$

$= 2(0.7781)$

$= 1.5562$

Find the value of $3^{2}-\log_{3}^{4}$

$3^{2} – \log_{3}^{4} = 3^{2} \log_{3} 3 – \log^{3} 4$

$= 3\log_{3} 3^{2} – \log_{3}^{4}$

$= 3\log_{3} 9 – \log_{3} 4$

$= 3\log_{3} (9/4)$
If

$\log M = – 2, 1423$ then to get its characteristic and mantissa is made positive as following

$=\frac{9}{4}=2\frac{1}{4}(\because a^{\log_{a}x}=x)$

If $\log 2 = 0.3010, \log 3 = 0.4771, \log 7 = 0.8451$ and $\log 11 = 1.0414$ , then find the value of the following :
$\log\frac{42}{11}$

$\log\frac{42}{11}=\log42-\log11$

$= \log(2 \times 3 \times 7) – \log 11$

$= \log 2 + \log 3 + \log 7 – \log 11$

$= 0.3010 + 0.4771 + 0.8451 – 1.0414$

$= 1.6232 – 1.0414$

$= 0.5818$

If $\log 2 = 0.3010, \log 3 = 0.4771, \log 7 = 0.8451$ and $\log 11 = 1.0414$ , then find the value of the following :
$\log5^{1/3}$

$\log5^{1/3}=\dfrac{1}{3}\log5$

$=\dfrac{1}{3}\log\dfrac{10}{2}=\dfrac{1}{3}(\log10-\log2)$

$=\dfrac{1}{3}(1-0.3010)=\dfrac{1}{3}(0.6990)$

$=0.2330$

Prove that:
$\log 10 + \log 100 + \log 1000 + \log 10000 = 10$

$L.H.S.$

$= \log 10 + \log 10^{2} + \log 10^{3} + \log 10^{4}$

$= \log 10 + 2 \log 10 + 3 \log 10 + 4 \log 10$

$= 10 \log 10$

$= 10 \times 1 (\because \log 10 = 1)$

$= 10$

$= R.H.S.$
Hence Proved.

If $\log 2 = 0.3010, \log 3 = 0.4771, \log 7 = 0.8451$ and $\log 11 = 1.0414$ , then find the value of the following :
$\log\frac{121}{120}$

$\log\dfrac{121}{120}=\log\dfrac{11*11}{12*10}$

$= \log 11^{2} – \log( 12 \times 10)$

$= 2 \log 11 – \log 12 – \log 10$

$= 2(1.0414)- log (2 \times 2 \times 3) – 1$

$= 2.0828 – 1 – 2 \log 2 – \log 3$

$= 1.0828 – 2(0.3010) – 0.4771$

$= 1.0828 – 0.6020 – 0.4771$

$= 1.0828 – 1.0791$

$= 0.0037$

Prove that:
$\log_{5} 3 . \log_{3} 4 . \log_{2} 5 = 2$

$L.H.S. = \log_{5}3 . \log_{3} 4. \log_{2} 5$

$=\dfrac{\log_{10}3}{\log_{10}5}*\dfrac{\log_{10}4}{\log_{10}3}*\dfrac{\log_{10}5}{\log_{10}2}\left ( \because \log_{b}n=\dfrac{\log_{a}n}{\log_{a}b} \right )$

$=\dfrac{\log_{10}4}{\log_{10}2}=\dfrac{2\log_{10}2}{\log_{10}2}=2=R.H.S.$

Find the logarithm of the following numbers by using log table :
$2813$

$2813$
Characteristic : There are characteristic will be

$4 – 1 = 3$ .
Mantissa : In log table, in first column, in front of 28 and under the column 1.
Find the number which is

$= 4487$ .
For fourth digit 3’s mean difference

$= 5$
On adding

$= 4492$
So, mantissa of

$\log_{10} 2813 = 0.4492$
Thus,

$\log_{10} 2813 =$ Characteristic + Mantissa

$= 3 + 0.4492 = 3.4492$

Find the characteristic of logarithm of following numbers :
$1270$

Number

$1270$ is

$4$ digit number.
So, characteristic of its logarithm will be

$4 – 1 = 3$ .

Find the logarithm of the following numbers by using log table :
$27.28$

$27.28$
Characteristic : Here, integral part is

$27$ which contains

$2$ digit. So, characteristic of its logarithm will be

$2 –1 = 1$ .
Mantissa : In log table, in first column, in front of 27 (ignoring decimal point) and under the column 2, find the number which is

$= 4346$
For fourth digit 8’s mean difference

$= 13$
On adding

$= 4359$
So, mantissa of

$\log_{10} 27.28 = 0.4359$
Thus,

$\log_{10} 27.28 =$ Characteristic + Mantissa

$= 1+ 0.04359 =1.4359$ .

Find the logarithm of the following numbers by using log table :
$0.678$

$0.678$
Characteristic : In

$0.678$ , there are no zero between decimal point and first significant digit. So, characteristic of its logarithm will be

$– (0+ 1) = – 1 = 1$
Mantissa : In log table, in first column, in front of 67 (ignoring decimal point) and under the column 8 find the number which is

$8312$ .
So, mantissa of

$\log_{10} 0.678 = 0.8312$
Thus,

$\log_{10} 0.678 =$ Characteristic + Mantissa

$=\bar{1} + 0.8312 = \bar{1}. 8312$

Find the logarithm of the following numbers by using log table :
$400$

$400$
Characteristic : There are 3 digit in number

$400$ . So, characteristic will be

$3 – 1 = 2$ .
Mantissa : In log table, in first column, in front of 40 and under the column 0, find the number which is

$6021$ .
So, mantissa of

$\log_{10} 400 = 0.6021$
Thus,

$\log_{10} 400 =$ Characteristic + Mantissa

$= 2 + 0.6021 =2.6021$ .

Give the solution of following questions in one term :
$\log 2 + 1$

$\log 2 + 1$

$= \log 2 + \log 10 (\because \log_{10} 10 = 1)$

$= \log (2 \times 10)$

$= \log 20$

Find the characteristic of logarithm of following numbers :
$20.125$

$20.125$ , integral part is

$20$ which contains

$2$ digit.
So, characteristic of its logarithm will be

$2 – 1 = 1$ .

Find the logarithm of the following numbers by using log table :
$9$

$9$
Characteristic : There are only

$1$ digit in number

$9$ . So, characteristic will be

$1 – 1 = 0$ .
Mantissa : In log table, in first column, in front of 90 and under the column 0 find the number which is =

$9542$ .
So mantissa of

$\log_{10} 9 = 0.9542$
Thus,

$\log_{10} 9 =$ Characteristic + Mantissa

$= 0 + 0.9542 = 0.9542$

Find the characteristic of logarithm of following numbers :
$70$

Number

$70$ is

$2$ digit number.
So, characteristic of its logarithm will be

$2 – 1 = 1$ .

Find the value of $x$ in the following :
$\log x = 0.452$

$\log x = 0.452$
Taking antilog on both sides
antilog

$(\log x)$ = antilog

$(0.452)$

$\Rightarrow x =$ antilog

$(0.452)$
(a) Mantissa of given number is

$0.452$ .
(b) In antilog table, common number of row of

$0.45$ and column of

$2$ , is

$2831$ .
(c) Characteristic of given number is 0. So, antilog of number will be l digit number.
(d) The number obtained in step (b) will be written as

$2.831$ . Thus antilog

$0.452 = 2.831$
Hence,

$x = 2.831$

Find the logarithm of the following numbers by using log table :
$0.08403$

$0.08403$
Characteristic : In

$0.08403$ , there are

$1$ zero between decimal point and first significant digit. So,
characteristic of its logarithm will be

$-(1 + 1) = – 2 = \bar{2}$
Mantissa

: In log table, in first column, in front of 84 (ignoring decimal

point) and under the column 0 find the number, which is

$= 9243$
For fourth digit 3 mean difference

$= \bar{2}$
On adding

$= 9245$
So, mantissa of

$\log_{10} 0.08403 = 0.9245$
Thus,

$\log_{10} 0.08403 =$ Characteristic + Mantissa

$= \bar{2}+ 0.9245 = \bar{2}.9245$ .

Find the logarithm of the following numbers by using log table :
$0.00003258$

$0.00003258$
Characteristic: In

$0.00003258$ , there are

$4$ zero between decimal point and first significant digit. So
characteristic of its logarithm will be

$-(4 + 1) = -5 = \bar{5}$
Mantissa : In log table, in first column, in front of 32 (ignoring decimal point) and under the column 5 find the number, which is

$= 5119$
For fourth digit

$8$ mean difference

$= 11$
On adding

$= 5130$
So, mantissa of

$log_{10} 0.00003258 = 0-5130$
Thus,

$\log_{10} 0.00003258 =$ Characteristic + Mantissa

$= \bar{5}+ 0.5130= \bar{5}.5130$

Find the logarithm of the following numbers by using log table :
$0.000287$

$0.000287$
Characteristic: In

$0.000287$ , there are

$3$ zero between decimal point and first significant digit. So,
characteristic of its logarithm will be

$-(3 + 1) = -4 =\bar{4}$
Mantissa : In log table, in first column, in front of 28 (ignoring decimal point) and under the column 7 find the number which is

$4579$ .
So, mantissa of

$\log_{10} 0.000287 = 0.4579$
Thus,

$\log_{10} 0.000287 =$ Characteristic + Mantissa

$=\bar{4} + 0.4579 = \bar{4}.4579$

Find the value of $x$ in the following :
$\log x = \bar{2}.6727$

$\log x = \bar{2}.6727$
Taking anitlog on both sides
antilog

$(\log x)$ = antilog

$\bar{2}.6727$

$\rightarrow x =$ antilog

$\bar{2}.6727$
(a) Mantissa of given number is

$0.6727$ .
(b) In antilog table, common number of row of

$0.67$ and column of

$2$ , is

$4699$ .
(c) In the same line, in the mean difference column of lead 7, number is 8.
(d) Sum of step (b) and (c)

$= 4699 + 8 = 4707$ .
(e) Characteristic of given number is

$\bar{2}$ . Thus

$2 – 1 = 1$ .
(f) The number obtained in step (d) will be written by one zero of right side as

$0.04707$ . Thus antilog

$\bar{2}.6727 = 0.4707$ .

Find the logarithm of the following numbers by using log table :
$0.00003208$

$0.00003208$
Characteristic: In

$0.00003208$ , there are

$4$ zero between decimal point and first significant digit. So,
characteristic of its logarithm will be

$-(4 + 1) = -5 = \bar{5}$
Mantissa : In log table, in first column, in front of 32 (ignoring decimal point) and under the column 0 find the number which is

$= 5051$
For fourth digit 8’s mean difference

$= 11$
On adding

$= 5062$
So, mantissa of

$\log_{10} 0.0003208 = 0.5062$
Thus,

$\log_{10} 0.0003208 =$ Characteristic + Mantissa

$= \bar{5}+ 0.5062$

$= \bar{5}.5062$

Find the logarithm of the following numbers by using log table :
$0.000125$

$0.000125$
Characteristic : In

$0.000125$ , there are

$3$ zero between decimal point and first significant digit. So,
characteristic of its logarithm will be

$– (3 + 1) = – 4 =\bar{4}$
Mantissa : In log table, in first column, in front of 12 (ignoring decimal point) and under the column 5 find the number which is

$= 0969$
So, mantissa of

$\log_{10} 0.000125 = 0.0969$
Thus,

$\log_{10}0.000125 =$ Characteristic + Mantissa

$= \bar{4}+ 0.0969 = \bar{4}.0969$

Find the logarithm of the following numbers by using log table :
$1.234$

$1.234$
Characteristic

$= 1- 1 = 0$
Mantissa

$= 0.0899 + 14 = 0.0913$

$\log_{10} 1.234 =$ Characteristic + Mantissa

$= 0 + 0.0913 = 0.0913$ .

Find the logarithm of the following numbers by using log table :
$0.0035$

$0.0035$
Characteristic : In

$0.0035$ , there are

$2$ zero between decimal point and first significant digit. So,
characteristic of its logarithm will be

$-(2+1) = -3 =\bar{3}$
Mantissa : In log table, in first column, in front of 35 (ignoring decimal point) and under the column 0, find the number which is

$= 5441$
So, mantissa of

$log_{10} 0.0035 = 0.5441$
Thus,

$\log_{10} 0.0035 =$ Characteristic + Mantissa

$=\bar{3} + 0.5441 = \bar{3}.5441$

If $\frac{\log144}{\log12}=\log x$ , then find the value of $x$ .

$\log x=\frac{\log 144}{\log 12}=\frac{\log (12)^{2}}{\log 12}=2*\frac{\log12}{\log12}$
On comparing,

$x = 100$
Hence,

$x=100$

Find the value of $\log 0.001$ .

$\log0.001=\log(\frac{0.001}{1})=\log(\frac{1}{1000})$

$= \log (10^{-3}) = -3 \log 10$

$= -3 * 1 = -3$ or

$3$

Prove that:
$\log_{3} 4 . \log_{4} 5 . \log_{5} 6 . \log_{6} 7 . \log_{7}\log_{8} 9 = 2$

$L.H.S.$

$= (\log_{3} 4 – \log_{4} 5)- (\log_{5} 6 . \log_{6} 7) (\log_{7} 8- \log_{8} 9)$

$= (\log_{3} 5 * \log_{5} 7) * \log_{7} 9$

$= \log_{3} 7 * \log_{7} 9$

$= \log_{3} 9 = log_{3} 3^{2}$

$= 2 * 1= R.H.S.$ Hence Proved.

If $\log 52.04 = 1.7163, \log 80.65 = 1.9066$ and $\log 9.753 = 0.9891$ , then find the value of
$\log\frac{52.04*80.65}{9.753}$

Given,

$\log 52.04= 1.7163$

$\log 80.65 = 1.9066$

$\log 9.753 = 0.9891$

$\log\frac{52.04*80.65}{9.753}$

$= \log (52.04 * 80.65) – \log 9.753$

$= \log 52.04 + \log 80.65 – \log 9.735$

$= 1.7163 + 1.9066-0.9891$

$= 3.6229 – 0.9891$

$=2.6338$
Hence,

$\log\frac{52.04*80.65}{9.753}=2.6338$

Prove that:
$\log_{10} \tan 1^{0}. \log_{10} \tan 2^{0}. \log_{10} \tan 89^{0} = 0$

$L.H.S.$

$= \log_{10} \tan 1^{0}. \log_{10} \tan 2^{0}…. \log_{10} \tan 45^{0} .....\log_{10} \tan 89^{0}$

$= \log_{10} \tan 1^{0}. \log_{10} \tan 2^{0}…. \log (1)…. \log_{10} \tan 89^{0}$

$= \log_{10} \tan 1^{0}- \log_{10} \tan 2^{0}…. \times 0 \times…. \log_{10} \tan 89^{0}$

$= 0$

$= R.H.S.$ Hence Proved.

If $\log 7 = 0.8451$ and $\log 3 = 0.4771$ , then find $\log (21)^{5}$ .

Given,

$\log 7 = 0.8451$ and

$\log 3 = 0.4771$

$\log (21)^{5} = \log (3 \times 7)^{5}$

$= 5 \log (3 \times 7)$

$= 5 \log 3 + 5 \log 7$

$= 5 \times 0.4771 + 5 \times 0.8451$

$= 2.3855 + 4.2255$

$= 6.6110$
Hence,

$\log (21)^{5} = 6.6110$

If $\log 2 = 0.3010$ , then find the value of $\log 200$ .

$\log 2 = 0.3010$ ……… (i)

$\log 200= \log (2 * 10^{2})$

$= \log 2 + \log 10^{2}$

$= \log 2 + 2 \log 10$

$= 0.3010 + 2 * 1$

$= 2 + 0.3010$

$= 2.3010$
Hence,

$\log 200 = 2.3010$

$\log 2 = 0.3010$ and $\log 3 = 0.4771$ , then find the value of $\log (0.06)^{6}$

Given,

$\log 2 = 0.3010$ and

$\log 3 = 0.4771$

$\log(0.06)^{6}=\log(\frac{6}{100})^{6}$

$=\log (6 * 10^{-2} )^{6}$

$= 6 \log (2 * 3 * 10^{-2} )$

$= 6 \log 2 + 6 \log 3 – 12 \log 10$

$= 6 * 0.3010 + 6 * 0.4771 – 12 * 1$

$= 1.8060 + 2.8626 – 12$

$= 4.6686 – 12 = -7.3314$

$= – 8 + 0.6686$

$= \bar{8}+ 0.6686 = \bar{8}.6686$
Hence,

$\log (0.06)^{6} = \bar{8}.6686$ or

$-7.3314$

If $\log 32.9= 1.5172, \log 568.1 = 2.7544$ and $\log 13.28 = 1.1232$ , then find the value of
$\log\frac{(13.28)^{3}}{32.9*568.1}$

Given,

$\log 32.9 = 1.5172, \log 568.1 =2.7544$ and

$\log 13.28 = 1.1232$

$\log\dfrac{(13.28)^{3}}{(32.9\times568.1)}$

$= \log (13.28)^{3} – \log (32.9 \times 568.1)$

$= 3 \log 13.28 – \log 32.9 – \log 568.1$

$= 3 \times 1.1232 – 1.5172-2.7544$

$= 3.3696-4.2716$

$= – 0.9020 = – 1 + 0.0980$

$= 1 + 0.098 = 1.0980$

$=\bar{1}+0.098=\bar{1}.0980$
Hence,

$\log\dfrac{(13.28)^{3}}{(32.9*568.1)}=-0.9020$ or

$\bar{1}0.980$

By using logarithm, find the value of
$\frac{520.4*8.065}{97.53}$

Let

$\frac{520.4\times8.065}{97.53}= x$
Taking

$\log$ on both sides,

$\log\left ( \frac{520.4\times8.065}{97.53} \right )=\log x$

$\log (520.4 \times 8.065) – \log 97.53 = \log x$

$\log 520.4 + \log 8.065 – \log 97.53 = \log x$ …..(i)

$2.7163 + 0.9066 – 1.9881 = \log x$

$3.6229 – 1.9881 = \log x$

$1.6348 = \log x$

$\therefore x =$ antilog

$1.6348 = 43.13$

$\log_{10} 3= 0.4771$ , then find $\log_{10} 0.027$ .

$\log_{10} 3 = 0.4771$

$\log_{10} 0.027 = log(\frac{27}{1000})$

$= \log_{10} (27 * 10^{-3})$

$= \log_{10} (3^{3} * 10^{-3} )$

$= \log_{10} (3)^{3} + \log_{10} (10)^{-3}$

$= 3 \log_{10} 3 – 3 \log_{10} 10$

$= 3 * 0.4771 – 3 * 1$

$= 1.4313 – 3$

$= – 1.5687$

$= -2 + 0.4313$

$= \bar{2}+ 0.4313 = \bar{2}.4313$
Hence,

$\log 0.027 = \bar{2}.4313$ or

$– 1.5687$

If $5^{-p} = 4^{-q} = 20^r$ ; show that:
$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0$

$5^{-p}=4^{-q}=20^{r}=k$

$\Rightarrow \: \log_{5}5^{-p}=\log_{5}k\:\:\Rightarrow \:\:p=-\log_{5}k$

Similarly,

$\:\: q=-\log_{4}k$ and

$\:\:r=\log_{20}k$

$\therefore \: \dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}=\dfrac{1}{-\log_{5}k}+\dfrac{1}{-\log_{4}k}+\dfrac{1}{\log_{20}k}$

$=\log_{k}\left (\dfrac{1}{5} \right )+\log_{k}\left (\dfrac{1}{4} \right )+\log_{k}20$

$=\log_{k}1=0$

$\log_x (\log_9 (3^x - 9)) < 1$

$3^{x}-9>0\:\:\Rightarrow \:\:x>2$

$\log_{x}\left ( \log_{9}\left ( 3^{x}-9 \right ) \right )<1$

$\Rightarrow \: \log_{9}\left ( 3^{x}-9 \right )<x^{1}=x\:\:\:(\because\: x>1)$

$\Rightarrow \:3^{x}-9<9^{x}$

Let

$3^{x}=t$

$\Rightarrow \: 3^{x}-9<\left ( 3^{2} \right )^{x}$

$\Rightarrow \: t-9<t^{2}\:\:\Rightarrow \:\:t^{2}-t+9>0$

$\rightarrow$ True for all

$t$ . Since discriminant of quadratic is always

$>0.$

$\Rightarrow \: x\:\epsilon \:(2,\infty)$

If $\dfrac {241}{4000}=\dfrac {241}{{2}^{m}{5}^{n}}$ , find $m$ and $n$ values?

It is given that,

$\cfrac{{241}}{{4000}} = \cfrac{{241}}{{{2^m}{5^n}}}$

Since, $4000 = {2^5} \times {5^3}$ , then,

$\cfrac{{241}}{{{2^5}{5^3}}} = \cfrac{{241}}{{{2^m}{5^n}}}$

By comparing both sides,

$m = 5$ and $n = 3$ .

Evaluate = $\sqrt[100]{10^{10^{10}}}$

$\sqrt [ 100 ]{ { 10 }^{ { 10 }^{ 10 } } } \\ ={ \left( { 10 }^{ { 10 }^{ 10 } } \right) }^{ \frac { 1 }{ 100 } }\\ ={ 10 }^{ { 10 }^{ 10 }\times \frac { 1 }{ 100 } }\\ ={ 10 }^{ { 10 }^{ 10 }\times \frac { 1 }{ { 10 }^{ 2 } } }\\ ={ 10 }^{ { 10 }^{ 10-2 } }\\ ={ 10 }^{ { 10 }^{ 8 } }\\ ={ 10 }^{ 100000000 }\\ =100.....100000000\quad$ zeros

Simplify the following expressions.
(a) $\displaystyle\, 5^{log \, 5/log \, 25}$

(b) $\displaystyle\, log_2 \, log \, 100$

$(a) 5^{(\log5/\log{25})}$

$=5^{\log_{25}5}=5^{1/2}=\sqrt{5}$

$(b)\log_2\log100$

$=\log_22$

$=1$

Simplify the following expressions.
$\displaystyle\, 0.(1 \, + \, 9^{log_3 \, 8})^{log_5 \, 5}$

$0.8.(1+9^{\log_38})^{\log_55}$

$=0.8.(1+3^{2\log_38})^1$

$=0.8.(1+3^{\log_3{64}})$

$=0.8(65)$

$=52$

Calculate $\displaystyle \log_5 9.8$ in terms of $a$ and $b$ , if $\log 2 = a$ and $\log 7 = b$ .

$\log_{5} 9.8 = \dfrac {\log 9.8}{\log 5} = \dfrac {\log \dfrac {98}{10}}{\log \dfrac {10}{2}} = \dfrac {\log 98 - \log 10}{\log 10 - \log 2}$

$= \dfrac {\log (49 \times 2) - 1}{1 - \log 2}$

Given:

$\log 2 = a$ and

$\log 7 = b$

$\Rightarrow \log_{5} 9.8 = \dfrac {(\log 7^{2}) + (\log 2) - 1}{1 - \log 2}$

$= \dfrac {2\log 7 + \log 2 - 1}{1 - \log 2}$

$= \dfrac {2b + a - 1}{1 - a}$ .

${x^{3{{\log }^3}{{10}^x} - {2 \over 3}\log {{10}^x} = 100\root 3 \of {10} }}$

1317131_1008967_ans_8fdf5918aa8e4e58a676b8cbf479e788.jpg

Solve the following equations.
$\displaystyle log_3 \, log_4 \, log_2 \, x \, = \, 0.$

$\log_3\log_4\log_2x=0$

$\implies \log_4\log_2x=1$

$\implies \log_2x=4^1=4$

$\implies x=2^4=16$

$\implies x=16$

Solve the following equation.
$\displaystyle\, \log^2_{1/2}\, (4x) + \log_2 \left ( \frac{x^2}{8} \right ) = 8$

${ \quad log }_{ \cfrac { 1 }{ 2 } }^2(4x)+{ log }_{ 2 }(\cfrac { { x }^{ 2 } }{ 8 } )=8\\ \Rightarrow (-{ log }_{ 2 }4x)^{ 2 }+{ log }_{ 2 }{ x }^{ 2 }-log_{ 2 }8=8\\ \Rightarrow { (2+{ log }_{ 2 }x) }^{ 2 }+{ 2log }_{ 2 }x-3=8$

Let,

${ log }_{ 2 }x=a$

Therefore,

$({ 2+a) }^{ 2 }+2a=11\\ \Rightarrow { a }^{ 2 }+6a-7=0\\ \Rightarrow a=-7,1$

Therefore,

${ log }_{ 2 }x=-7,1$

$x={ 2 }^{ -7 },2.$

Solve $\sqrt{\log_2\ x^{4}}+4\log_4\sqrt{\dfrac{2}{x}}=2$

$\sqrt{\log_{2}{{x}^{4}}}+4\log_{4}{\sqrt{\dfrac{2}{x}}}=2$

$\Rightarrow\,\sqrt{4\log_{2}{x}}+4\log_{4}{\sqrt{2}}-4\log_{4}{\sqrt{x}}=2$

$\Rightarrow\,2\sqrt{\log_{2}{x}}+4\log_{4}{\sqrt{2}}-4\log_{4}{\sqrt{x}}=2$

$\Rightarrow\,\sqrt{\log_{2}{x}}+2\log_{4}{\sqrt{2}}-2\log_{4}{\sqrt{x}}=1$

$\Rightarrow\,\sqrt{\log_{2}{x}}+\dfrac{1}{2}\log_{2}{2}-\dfrac{2}{2}\log_{2}{x}=1$

$\Rightarrow\,\sqrt{\log_{2}{x}}+\dfrac{1}{2}-\dfrac{2}{2}\log_{2}{x}=1$

$\Rightarrow\,\sqrt{\log_{2}{x}}-\dfrac{2}{2}\log_{2}{x}=1-\dfrac{1}{2}$

$\Rightarrow\,\sqrt{\log_{2}{x}}-\log_{2}{x}=\dfrac{1}{2}$

Let

$\alpha=\sqrt{\log_{2}{x}}\Rightarrow\,{\alpha}^{2}=\log_{2}{x}$

$\Rightarrow\,\alpha-{\alpha}^{2}=\dfrac{1}{2}$

$\Rightarrow\,2\alpha-2{\alpha}^{2}-1=0$

$\Rightarrow\,2{\alpha}^{2}-2\alpha-1=0$

$\Rightarrow\,\alpha=\dfrac{2\pm\sqrt{4+8}}{4}=\dfrac{2\pm\,2\sqrt{3}}{4}=\dfrac{\pm\,\sqrt{3}}{2}$

$\Rightarrow\,\sqrt{\log_{2}{x}}=\dfrac{1+\,\sqrt{3}}{2}\,\,\,\,\,\,\,\,\because\sqrt{x}>0$

$\Rightarrow\,\log_{2}{x}={\left(\dfrac{1+\,\sqrt{3}}{2}\right)}^{2}=\dfrac{1+3+2\sqrt{3}}{4}=\dfrac{2+\sqrt{3}}{2}$

$\therefore\,x={2}^{\frac{2+\sqrt{3}}{2}}$

$\log _{ x }{ y } =10\log _{ 2 }{ y } =100$ then $y=$

$\log_{x}{y}=10\Rightarrow {x}^{10}=y$ .....

$\left(1\right)$

$\log_{2}{y}=100\Rightarrow y={2}^{100}$ .....

$\left(2\right)$

From

$\left(1\right)$ and

$\left(2\right)$ we have

${x}^{10}={2}^{100}$

$\Rightarrow {x}^{10}={\left({2}^{10}\right)}^{10}$

$\Rightarrow x={\left({\left({2}^{10}\right)}^{10}\right)}^{\frac{1}{10}}$ since

${\left({a}^{m}\right)}^{n}={a}^{mn}$

$\Rightarrow x={2}^{10}=1024$

$\Rightarrow y={x}^{10}$ from

$\left(1\right)$

$\Rightarrow y={\left({2}^{10}\right)}^{10}={2}^{100}$ since

${\left({a}^{m}\right)}^{n}={a}^{mn}$

$\therefore y={2}^{100}$

Find approximate value of ${\log _e}(4.04)\,\ \ if \ \ \,{\log _{10}}4\, = \,0.6021\,and\,{\log _{10}}e = 0.4343$

1188497_1175685_ans_c9d7c26987ec41bf9d45f149f4e5775f.jpg

Solve for $x$
${({a^{3x + 5}})^2}\times {({a^x})^4} = {a^{8x + 12}}$

1321167_1079448_ans_877188e1fefa4a62bdb4604241118d82.jpeg

Prove that :
$\dfrac{2^{30} + 2^{29}+ 2^{28} }{2^{31} + 2^{30}- 2^{29}}=\dfrac{7}{10}$

$\dfrac { { 2 }^{ 30 }+{ 2 }^{ 29 }+{ 2 }^{ 28 } }{ { 2 }^{ 31 }+{ 2 }^{ 30 }-{ 2 }^{ 29 } } =\dfrac { { { 2 }^{ 28 }(2 }^{ 2 }+{ 2 }^{ 1 }+1) }{ { { 2 }^{ 29 }(2 }^{ 2 }+{ 2 }^{ 1 }-1) }$

$=\dfrac { { { 2 }^{ 28 }(4 }+{ 2 }+1) }{ { { 2 }^{ 29 }(4 }+{ 2 }-1) }$

$=\dfrac { { { 2 }^{ 28 }(7 }) }{ { { 2 }^{ 29 }(5 }) }$

applying

$\dfrac { { a }^{ m } }{ { a }^{ n } } ={ a }^{ m.n }$

$\dfrac { { { 2 }^{ 28-29 }(7 }) }{ 5 } =\dfrac { { { 2 }^{ -1 }(7 }) }{ 5 }$

${ a }^{ -n }=\dfrac { 1 }{ { a }^{ n } } =\dfrac { 7 }{ 10 }$

Given: $2 \log _ { 10 ^ { x } } + 1 = \log _ { 10 } 250$ . Find $x$ and $\log _ { 10 } 2 x$

$2\log _{ 10 }{ x } +\log _{ 10 }{ 10 } =\log _{ 10 }{ 250 }$

$=\log _{ 10 }{ { x }^{ 2 } } +\log _{ 10 }{ 10 } =\log _{ 10 }{ 250 }$

$\therefore 10{ x }^{ 2 }=250\Rightarrow x=5$

$\log _{ 10 }{ 2x } =\log _{ 10 }{ 10 } =1$

Find the value of $\log { _{ 5 } }$ $0.008$

$\log_5 0.008=x$

or,

$0.008=5^x$

or,

$5^x =\dfrac {8}{1000}$

or,

$5^x =\dfrac {1}{125} =\dfrac {1}{5^3}$

or,

$5^x =5^{-3}$

$\therefore \ x=-3$

$\therefore \ \log_5 0.008=-3$

log 6 + 2 log 5 + log 4 - log 3 - log 2

$\textbf{Step 1: Use the properties of log}$

$\log 6 + 2\log 5 + \log 4 - \log 3 - \log 2$

$= \log 6 + 2\log 5 + { \log 2 }^{ 2 }-\left( \log 3 + \log 2 \right)$ [

$\boldsymbol {\because \log m + \log n = \log ( m\times n)}$ ]

$= \log 6 + 2\log 5 + 2\log 2 - \log\left( 3\times 2 \right)$

$= \log 6 + 2\left( \log 5 + \log 2 \right) - \log 6$

$= \log 6 + 2\log\left( 5\times 2 \right) - \log 6$

$=2\log 10$

$=2$ [

$\boldsymbol {\because \log 10 = 1 }$ ]

$\textbf{Hence , } \boldsymbol{ \log 6 + 2 \log 5 + \log 4 - \log 3 - \log 2 } \textbf{ is equal to 2 }$

Solve:
$\dfrac{1}{1+a^{n-m}}+\dfrac{1}{1+a^{m-n}}$

$\begin{array}{l} \dfrac { 1 }{ { 1+{ a^{ \left( { n-m } \right) } } } } +\dfrac { 1 }{ { 1+{ a^{ -\left( { n-m } \right) } } } } \\ \Rightarrow \dfrac { 1 }{ { 1+{ a^{ \left( { n-m } \right) } } } } +\dfrac { 1 }{ { 1+\dfrac { 1 }{ { { a^{ \left( { n-m } \right) } } } } } } \\ \Rightarrow \dfrac { 1 }{ { 1+{ a^{ \left( { n-m } \right) } } } } +\dfrac { { { a^{ \left( { n-m } \right) } } } }{ { 1+{ a^{ \left( { n-m } \right) } } } } \\ \dfrac { { 1+{ a^{ \left( { n-m } \right) } } } }{ { 1+{ a^{ \left( { n-m } \right) } } } } =1\, \, \, Ans. \end{array}$

Express in terms of bases to the power of exponenets
$8.9^{2}$

${8.9^2} = {2^3}.{\left( {{3^2}} \right)^2} = {2^3}{.3^4}$

Evaluate : $\left( \dfrac { 5 } { 3 } \right) ^ { x } \cdot \left( \dfrac { 9 } { 25 } \right) ^ { x ^ { 2 } + 2 x - 11 } = \left( \dfrac { 5 } { 3 } \right) ^ { 9 }$

$\left( \dfrac{5}{3}\right)^x.\left( \dfrac{9}{25}\right)^{x^2+2x-11}=\left( \dfrac{5}{3}\right)^9$

$\Rightarrow \left( \dfrac{5}{3}\right)^x.\left( \left( \dfrac{3}{5}\right)^2\right)^{x^2+2x+11}=\left( \dfrac{5}{3}\right)^9$

$\Rightarrow \left( \dfrac{5}{3}\right)^x.\left( \dfrac{5}{3}\right)^{-2\left( x^2+2x-11\right)}=\left( \dfrac{5}{3}\right)^9$

$\Rightarrow a^m.a^n=a^{m+n}$

$\Rightarrow \left( \dfrac{5}{3}\right)^{x+\left( -2\left( x^2+2x-11\right)\right)}=\left( \dfrac{5}{3}\right)^9$

comparing the powers,

$\Rightarrow x+\left( -2\left( x^2+2x-11\right)\right)=9$

$x+\left( -2x^2-4x+22\right)=9$

$x-2x^2-4x+22=9$

$-2x^2-3x+22-9=0$

$-2x^2-3x+13=0$

$\Rightarrow a=-2$

$b=-3$

$c=13$

$\Rightarrow D=b^2-4ac$

$=\left( -3\right)^2-4\times \left(-2\right)\times 13$

$=9+104$

$=113$

$\Rightarrow x=\dfrac{-b\pm \sqrt{D}}{2a}$

$=\dfrac{-\left(-3\right)\pm \sqrt{113}}{2\left( -2\right)}$

$=\dfrac{3\pm \sqrt{113}}{-4}$

$=\dfrac{-3\pm \sqrt{113}}{4}$

Hence, the answer is

$\dfrac{-3\pm \sqrt{113}}{4}.$

Solve:
$(-48p^{4})\div(-9p^{2})$

Given,

$(-48p^{4})\div(-9p^{2})$

We can write it as:

$\dfrac { { -48{ p^{ 4 } } } }{ { -9{ p^{ 2 } } } }$

$=\dfrac{48}{9}\times \dfrac{p^4}{p^2}$

We know that,

$\dfrac{a^m}{a^n}=a^{m-n}$

Hence,

$\dfrac{48}{9}\times p^{4-2}$

$\dfrac{16}{3}\times p^2$

$=\dfrac{16p^2}{3}$

Hence,

$(-48p^{4})\div(-9p^{2})=\dfrac{16p^2}{3}$ .

the value $\left( \frac { 1 }{ \sqrt { 27 } } \right) ^{ 2-\left( { log }_{ 5 }{ 16/2log }_{ 5 }9 \right) } =$

2039448_1286150_ans_77edff051ee541458b6d1dfb32964c67.jpg

Find the appropriate value of $\log_{10}$ $(1016)$ , given $\log_{10}e=0.4343$

$Lo{ g_{ 10 } }\left( { 1016 } \right) =? \\ \log _{ 10 }{ e }=0.4343$

$\\ Let \\ f\left( x \right) =\log _{ 10 } { x }\to (i) \\ f\left( x \right) =\dfrac { { \log _{ e }{ x } } }{ { \log _{ e } { 10 } } } \to (ii) \\ =\log _{ e } { x }\cdot \log _{ 10 } { e } \\ =0.4343\log _{ e } { x } \\$

On differentiating w.r.t

$x$ , we get

$f'\left( x \right) =\dfrac { { 0.4343 } }{ x } \\$

Since,

$x=1016 \\ =1000+16 \\$

$Let\, \, h=16,a=1000 \\ f\left( a \right) =f\left( { 1000 } \right) \\ =\log _{ 10 } { 1000 } \\ =3\log _{ 10 } { 10 } \\ =3\times 1 \\= 3$

$\\ f'\left( a \right) =f'\left( { 1000 } \right) =\dfrac { { 0.4343 } }{ { 1000 } } \\ =0.0004343 \\ f\left( { a+h } \right) \approx f\left( a \right) +hf'\left( a \right) \\ \approx 3+1610.0004343 \\ k\approx 3+0.009488$

$\approx 3.0069488$

Hence, this is the answer.

$|x-1|^A = (x-1)^7$ , where $A=log_3x^2 -2log_x9.$

Domain of

$x=\left(0,\infty\right)-\left\{1\right\}$

${\left|x-1\right|}^{{\log}_{3}{{x}^{2}}-2{\log}_{x}{9}}={\left(x-1\right)}^{7}$

$\Rightarrow {\left(x-1\right)}^{2\left(\log_{3}{x}-{\log}_{x}{9}\right)}={\left(x-1\right)}^{7}$

$\Rightarrow 2\left(\log_{3}{x}-{\log}_{x}{9}\right)=7$

$\Rightarrow 2\left(\log_{3}{x}-2{\log}_{x}{3}\right)=7$

$\Rightarrow 2\left(\log_{3}{x}-\dfrac{2}{{\log}_{3}{x}}\right)=7$ sicne

$\log_{a}{x}=\dfrac{1}{\log_{x}{a}}$

Let

$\log_{3}{x}=t$

$\Rightarrow 2\left(t-\dfrac{2}{t}\right)=7$

$\Rightarrow 2\left(\dfrac{{t}^{2}-2}{t}\right)=7$

$\Rightarrow 2{t}^{2}-7t-4=0$

$\Rightarrow \left(2t+1\right)\left(t-4\right)=0$

$\Rightarrow t=\dfrac{-1}{2}$ ,

$t=4$

$\Rightarrow \log_{3}{x}=\dfrac{-1}{2}$ ,

$\log_{3}{x}=4$

$\Rightarrow x=\dfrac{1}{\sqrt{3}},81$

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log 14{^{.}4}.$

1573776_1743827_ans_cbbec83a376b4aaeb553a2c469bf3a6c.png

From the tables find the seventh root of $.000026751$ . Making use of the tables, find the approximate values.

1600012_1731046_ans_1aadf5aa513c4af2b088bab45633a1b5.jpg

Evaluate: $\log_{3}{2}\log_{4}{3}\log_{5}{4}...\log_{64}{63}$

Given,

$\log_{3}{2}\log_{4}{3}\log_{5}{4}...\log_{64}{63}$

$=\dfrac{\log{2}}{\log{3}}\times\dfrac{\log{3}}{\log{4}}\times\dfrac{\log{4}}{\log{5}}\times ...\times\dfrac{\log{63}}{\log{64}}$

$=\log_{64}{2}$

$=\log_{{2}^{6}}{2}$

$=\dfrac{1}{6}\log_{2}{2}$

$=\dfrac{1}{6}$

$log_2(4^x-5.2^x+2) > 2$ .

$\log_2(4^x-5.2^x+2)>2$

$=4^x-5.2^x+2 >2^2$

$=(2^x)^2-5.2^x+2>4$

Let

$2^x=t$

$t^2-5t-2>0$

$t=\cfrac {5\pm\sqrt {25+8}}{2}=\cfrac {5\pm\sqrt {33}}{2}$

$=\left(t-\left(\cfrac {5+\sqrt {33}}{2}\right)\right)\left(t-\left(\cfrac {5-\sqrt {33}}{2}\right)\right) >0$

(Refer to Image)

$\therefore$

$2^x\epsilon \underset {negative}{\left(-\infty,\cfrac {51-\sqrt {33}}{2}\right)}t\left(\cfrac {5+\sqrt {33}}{2},\infty\right)$

$\log$ is not defined

$\therefore$

$x\epsilon \left(\log_2\cfrac {5+\sqrt {33}}{2},\log_2\infty\right)$ .

1231124_1342238_ans_f06fd779affb4afdaaabe51e36882aed.JPG

Find, by inspection the characteristic of the logarithms of $21735, 23 ^{.}8, 350, {^{.}035}, {^{.}2}, {^{.}87}, {^{.}875}.$

1573769_1743787_ans_f7c766fbc2574b9ca82d9e19362a5392.png

$\log 2 ^ { 3 } + \log 3 + \log 5 = ?$

$\textbf{Step 1 : Use the addition properties of log}$

$\text{Given , } \log 2^3 + \log 3 + \log 5$

$= \log 8 + \log 3 + \log 5$

$= \log ( 8 \times 3 \times 5 )$ [

$\boldsymbol {\because \log m + \log n = \log (m\times n) }$ ]

$= \log 120$

$\textbf{Hence , } \boldsymbol {\log 2^3 + \log 3 + \log 5 } \textbf { is equal to } \boldsymbol {\log 120}$

Which is greater $x=\log _{ 3 }{ 5 } \quad or\quad y=\log _{ 17 }{ 25 }$ ?

$x=\log_{3}{5}=\log_{{3}^{2}}{{5}^{2}}=\log_{9}{25}=\dfrac{1}{\log_{25}{9}}$ using

$\log_{b}{a}=\dfrac{1}{\log_{a}{b}}$

$y=\log_{17}{25}=\dfrac{1}{\log_{25}{17}}$ using

$\log_{b}{a}=\dfrac{1}{\log_{a}{b}}$

We know that

$17>9$

$\log_{25}{17}>\log_{25}{9}$

$\dfrac{1}{\log_{25}{17}}<\dfrac{1}{\log_{25}{9}}$

$\Rightarrow \log_{17}{25}<\log_{9}{25}$

$\Rightarrow \log_{17}{25}<\dfrac{2}{2}\log_{3}{5}$ using

$\log_{{b}^{m}}{{a}^{n}}=\dfrac{n}{m}\dfrac{1}{\log_{a}{b}}$

$\therefore \log_{17}{25}<\log_{3}{5}$

$\therefore \log_{3}{5}>\log_{17}{25}$

Hence

$x>y$

Show that :
$1/ \log_{8} 36 + 1 / \log_{9} 36+ 1 / \log_{18} 36=2$

1974335_1811300_ans_84a84eaeab844e2cb02180f4c8f87d55.jpg

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log 4\dfrac{2}{3}.$

1573777_1743829_ans_fe1313a10e7643dabb8024f0f0838b4f.png

Find the value of $\log 6 + 2 \log 5 + \log 4 \log 3 \log 2$ .

$\textbf{Step 1: Use the properties of log}$

$\log 6 + 2\log 5 + \log 4 - \log 3 - \log 2$

$= \log 6 + 2\log 5 + { \log 2 }^{ 2 }-\left( \log 3 + \log 2 \right)$

$\boldsymbol {[\because \log m + \log n = \log ( m\times n)]}$

$= \log 6 + 2\log 5 + 2\log 2 - \log\left( 3\times 2 \right)$

$= \log 6 + 2\left( \log 5 + \log 2 \right) - \log 6$

$= \log 6 + 2\log\left( 5\times 2 \right) - \log 6$

$=2\log 10$

$=2$ [

$\boldsymbol {\because \log 10 = 1 }$ ]

$\textbf{Hence , } \boldsymbol{ \log 6 + 2 \log 5 + \log 4 - \log 3 - \log 2 } \textbf{ is equal to 2 }$

Logarithm And Antilogarithm - Class 11 Commerce Applied Mathematics - Extra Questions

Write the characteristic of 1235.3 by using their standard forms:

Simplify: 8−1×532−4\dfrac{8^{-1} \times 5 ^{3}}{2^{-4}}

If log10.001=a\displaystyle \log_{10} .001 = a, then −a-a=

Compute:(47)2×(4−3)4(4^7)^2\times (4^{-3})^4

Compute:(5628)0÷(25)3×1625(\dfrac{56}{28})^0 \div (\dfrac{2}{5})^3 \times \dfrac{16}{25}

Compute:(2^{-9}\div 2^{-11})^4 (2^{-9}\div 2^{-11})^4

Find the characteristic of the logarithm of the number 5395.5395.

(12)^{-2} \times 3^2 (12)^{-2} \times 3^2 is equal to \dfrac{1}{16} \dfrac{1}{16} .If true then enter 11 and if false then enter 00

Compute the absolute value of: (-5)^4 \times (-5)^6 \div (-5)^{9} (-5)^4 \times (-5)^6 \div (-5)^{9}

Compute:(9^0 \times 4^{1}) (9^0 \times 4^{1})

The value of x,if \log_{10} x = -2\log_{10} x = -2 is kk.then 100x100x=?

Show that \displaystyle \left ( 2\sqrt{2} \right )^{-2/3}= \frac{1}{2}.\displaystyle \left ( 2\sqrt{2} \right )^{-2/3}= \frac{1}{2}. can be written as \displaystyle \log _{2\sqrt{2}} 2= -\frac{2}{3}\displaystyle \log _{2\sqrt{2}} 2= -\frac{2}{3}

\displaystyle \log _{3}\frac{1}{243}= -5.\displaystyle \log _{3}\frac{1}{243}= -5.

Show that \displaystyle 5^{0}= 1\displaystyle 5^{0}= 1 can be written as \displaystyle \log _{25}1= 0\displaystyle \log _{25}1= 0

Express \displaystyle \log _{100}0.1= -\frac{1}{2}\displaystyle \log _{100}0.1= -\frac{1}{2} in exponential form:

\displaystyle 4^{3/2}= 8\displaystyle 4^{3/2}= 8

If 3^{-2}3^{-2} is \cfrac{1}{m}\cfrac{1}{m}, then the value of mm is

If (-4)^{-2}(-4)^{-2} is \cfrac{1}{m}\cfrac{1}{m}, then the value of mm is

Find the value of: (3^0 + 4^{-1}) \times 2^2(3^0 + 4^{-1}) \times 2^2

Evaluate \displaystyle \left ( \frac{5}{8} \right )^{-7} \times \left ( \frac{8}{5} \right )^{-4}\displaystyle \left ( \frac{5}{8} \right )^{-7} \times \left ( \frac{8}{5} \right )^{-4} is \frac { 512 }{m } \frac { 512 }{m } Value of mm is

Evaluate (5^{-1} \times 2^{-1}) \times 6^{-1}(5^{-1} \times 2^{-1}) \times 6^{-1} is \frac{1}{m}\frac{1}{m}Value of mm is

Evaluate \displaystyle \frac{8^{-1} \times 5^3}{2^{-4}}\displaystyle \frac{8^{-1} \times 5^3}{2^{-4}}

Find the value of mm for which 5^m \div 5^{-3} = 5^55^m \div 5^{-3} = 5^5

Evaluate: -\displaystyle \left \{ \left ( \frac{1}{3} \right )^{-1} - \left ( \frac{1}{4} \right )^{-1} \right \}^{-1}-\displaystyle \left \{ \left ( \frac{1}{3} \right )^{-1} - \left ( \frac{1}{4} \right )^{-1} \right \}^{-1}

Find xx, if (4^{-1} + 8 ^{-1} ) \times (3^{-1} - 9^{-1}) \div \displaystyle \frac{1}{12} = 5^x(4^{-1} + 8 ^{-1} ) \times (3^{-1} - 9^{-1}) \div \displaystyle \frac{1}{12} = 5^x

By what number should \displaystyle \left ( \frac{-2}{3} \right )^{-3}\displaystyle \left ( \frac{-2}{3} \right )^{-3} be divided so that the quotient may be \displaystyle \left ( \frac{4}{27} \right )^{-2}\displaystyle \left ( \frac{4}{27} \right )^{-2}

Simplify: \left(\cfrac{3}{2} \right)^{0} \times \left(\cfrac{4}{5}\right)^{-2}\left(\cfrac{3}{2} \right)^{0} \times \left(\cfrac{4}{5}\right)^{-2}

Simplify : \displaystyle \frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} (t \neq 0)\displaystyle \frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} (t \neq 0)

Simplify: 3^0 +2^{-2}3^0 +2^{-2}

Simplify \displaystyle \left [ \left ( \frac{2}{3} \right )^{-1} \times \left ( \frac{3}{4} \right )^{-1} \right ]^{-1}\displaystyle \left [ \left ( \frac{2}{3} \right )^{-1} \times \left ( \frac{3}{4} \right )^{-1} \right ]^{-1}

Find the value \displaystyle 27^{\cfrac{1}{3} } \times 16^{\cfrac{-1}{4} }\displaystyle 27^{\cfrac{1}{3} } \times 16^{\cfrac{-1}{4} }

By what number should \displaystyle \left ( \frac{3}{4} \right )^{-3}\displaystyle \left ( \frac{3}{4} \right )^{-3} be divided so that the quotient becomes 128128 ?

Find the value (512)^{\frac{-2}{9}} (512)^{\frac{-2}{9}}

For what negative values of xx, will x^{18}x^{18} be equal to x^{20}x^{20}?

For what positive values of xx, will x^{18}x^{18} be greater than x^{20}x^{20}?

For what negative values of xx, will x^{20}x^{20} be greater than x^{18}x^{18}?

If \displaystyle \log_{3}a=4\displaystyle \log_{3}a=4, then find the value of aa.

Find the value of x if x^3 = \displaystyle \left ( \frac{6}{5} \right )^{-3} \times \left ( \frac{6}{5} \right )^6.x^3 = \displaystyle \left ( \frac{6}{5} \right )^{-3} \times \left ( \frac{6}{5} \right )^6.

{ 3 }^{ 3 }\times { 3 }^{ 6 }\times { 3 }^{ 7 }={ 3 }^{ 3 }\times { 3 }^{ 6 }\times { 3 }^{ 7 }=?

Suppose mm and nn are distinct integers. Can \cfrac { { 3 }^{ m }\times { 2 }^{ n } }{ { 2 }^{ m }\times { 3 }^{ n } } \cfrac { { 3 }^{ m }\times { 2 }^{ n } }{ { 2 }^{ m }\times { 3 }^{ n } } be an integer? Give reasons.

{ 2 }^{ 5 }\times { 5 }^{ 2 }\times { 2 }^{ 3 }\times 5={ 2 }^{ 5 }\times { 5 }^{ 2 }\times { 2 }^{ 3 }\times 5=?

{ 2 }^{ 5 }\times { 2 }^{ 6 }=2^? { 2 }^{ 5 }\times { 2 }^{ 6 }=2^?

Simplify:{ 3 }^{ 1 }\times { 3 }^{ 2 }\times { 3 }^{ 3 }\times { 3 }^{ 4 }\times { 3 }^{ 5 }\times { 3 }^{ 6 }{ 3 }^{ 1 }\times { 3 }^{ 2 }\times { 3 }^{ 3 }\times { 3 }^{ 4 }\times { 3 }^{ 5 }\times { 3 }^{ 6 }

Simplify:{2}^{2}\times {3}^{3}\times {2}^{4}\times {3}^{5}\times {3}^{6}{2}^{2}\times {3}^{3}\times {2}^{4}\times {3}^{5}\times {3}^{6}

Simplify and give reasons:{ \left[ \left( { 3 }^{ 2 }-{ 2 }^{ 2 } \right) \div \cfrac { 1 }{ 5 } \right] }^{ 2 }{ \left[ \left( { 3 }^{ 2 }-{ 2 }^{ 2 } \right) \div \cfrac { 1 }{ 5 } \right] }^{ 2 }

Rakesh solved some problems of exponents in the following way. Do you agree with the solutions? If not why? Justify your argument.{ x }^{ -3 }\times { x }^{ -2 }={ x }^{ -6 }{ x }^{ -3 }\times { x }^{ -2 }={ x }^{ -6 }

If \log_{2}x = 3\log_{2}x = 3, then x =x = ______

Find the value of 'xx' such that\cfrac{1}{49}\times {7}^{2x}={7}^{8}\cfrac{1}{49}\times {7}^{2x}={7}^{8}

Using logarithmic table find the value of the following.(i) \log { 23.17 }\log { 23.17 }(ii) \log { 9.321 } \log { 9.321 }(iii) \log { 329.5 } \log { 329.5 } (iv) \log { 0.001364 } \log { 0.001364 } (v) \log { 0.9876 }\log { 0.9876 }(vi) \log { 6576 } \log { 6576 }

Find, by inspection, the characteristics of the logarithms of 21735, 23.8, 350, .035, .2, .87, .87521735, 23.8, 350, .035, .2, .87, .875.

Given \log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980\log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980, find the value of \log \sqrt [3]{12}\log \sqrt [3]{12}.

Given \log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980\log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980, find the value of \log 14.4\log 14.4.

Given \log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980}, find the value of \log 84\log 84.

Given \log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980\log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980, find the value of \log 4\dfrac{3}{2}\log 4\dfrac{3}{2}.

Given \log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980}, find the value of\log {^{.}128}.\log {^{.}128}.

Given \log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980}, find the value of \log 64\log 64.

Give the position of the first significant figure in the numbers whose logarithms are\bar{2} {^{.}7781513}, {^{.}6910815}, \bar{5} {^{.}4871384}.\bar{2} {^{.}7781513}, {^{.}6910815}, \bar{5} {^{.}4871384}.

How many digits are there in the integral part of the numbers whose logarithms are respectively4{^{.}30103}, 1{^{.}4771213}, 3{^{.}69897}, {^{.}56515}?4{^{.}30103}, 1{^{.}4771213}, 3{^{.}69897}, {^{.}56515}?

Given \log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980}, find the value of\log {^{.}0125}.\log {^{.}0125}.

Given \log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980}, find the value of\log \sqrt[4]{^{.}0105}.\log \sqrt[4]{^{.}0105}.

Find the product of 37.203, 3.7203, .037203, 37203037.203, 3.7203, .037203, 372030, having given that\log 37.203 = 1.5705780\log 37.203 = 1.5705780, and \log 1915631 = 6.2823120\log 1915631 = 6.2823120.

Evaluate : 2^5\times 2^8\div 2^62^5\times 2^8\div 2^6

Given \log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980}, find the value of\log \sqrt{\dfrac{35}{27}}.\log \sqrt{\dfrac{35}{27}}.

Solve the following equations.\displaystyle log_2 \, (4^x \, + \, 4) \, - \, log_2 \, (2{x \, + \, 1} \, - \, 3) \, = \, 0.\displaystyle log_2 \, (4^x \, + \, 4) \, - \, log_2 \, (2{x \, + \, 1} \, - \, 3) \, = \, 0.

Write value of \sqrt [ 3 ]{ 2 } \times \sqrt [ 4 ]{ 2 } \times \sqrt [ 12 ]{ 32 } \sqrt [ 3 ]{ 2 } \times \sqrt [ 4 ]{ 2 } \times \sqrt [ 12 ]{ 32 }

Solve the following equation:\displaystyle\, 5^{x - 1} = 10^x\cdot 2^{-x}\cdot 5^{x + 1}\displaystyle\, 5^{x - 1} = 10^x\cdot 2^{-x}\cdot 5^{x + 1}

Rewrite the following equation in the logarithm from :5^0 \, = \, 15^0 \, = \, 1

Rewrite the following equation in the logarithm from :(2\sqrt{2})^{-2/3} \, = \, \dfrac{1}{2}.(2\sqrt{2})^{-2/3} \, = \, \dfrac{1}{2}.

Rewrite the following equation in the exponential form.log_{5\sqrt{5}} \, 5 \, = \, \dfrac{2}{3}log_{5\sqrt{5}} \, 5 \, = \, \dfrac{2}{3}

Rewrite the following equation in the exponential form.log_{100} \, 0.1 \, = \, -\dfrac{1}{2}log_{100} \, 0.1 \, = \, -\dfrac{1}{2}

Solve x^{\sqrt{x}} \, = \, (\sqrt {x})^xx^{\sqrt{x}} \, = \, (\sqrt {x})^x

Find the value of {(23.17)^{{1 \over {5.76}}}}{(23.17)^{{1 \over {5.76}}}} using log table.

Rewrite the following equation in the exponential form.\log_3 \, \frac{1}{243} \, = \, -5.\log_3 \, \frac{1}{243} \, = \, -5.

Evaluate : {2^{{{\log }_3}5}} - {5^{{{\log }_3}2}}{2^{{{\log }_3}5}} - {5^{{{\log }_3}2}}

{ \left[ { \left( \dfrac { 15 }{ 12 } \right) }^{ 3 } \right] }^{ 4 }{ \left[ { \left( \dfrac { 15 }{ 12 } \right) }^{ 3 } \right] }^{ 4 }

If x=243x=243, then find the value of {x^{\frac{1}{5}}} \times {x^{ - \frac{1}{5}}}{x^{\frac{1}{5}}} \times {x^{ - \frac{1}{5}}}

Solve: \log (3x + 2) + \log (3x - 2) = \log5\log (3x + 2) + \log (3x - 2) = \log5

Convert the following to logarithmic form:5^{2} =255^{2} =25

If X^\cfrac{a}{b} =1, X^\cfrac{a}{b} =1, then find the value of 'a''a'.

Simplify:

$\dfrac{8^{-1} \times 5 ^{3}}{2^{-4}}$

If $\displaystyle \log_{10} .001 = a$ , then $-a$ =

Compute:
$(4^7)^2\times (4^{-3})^4$

Compute:
$(\dfrac{56}{28})^0 \div (\dfrac{2}{5})^3 \times \dfrac{16}{25}$

Compute:
$(2^{-9}\div 2^{-11})^4$

Find the characteristic of the logarithm of the number $5395.$

$(12)^{-2} \times 3^2$ is equal to $\dfrac{1}{16}$ .
If true then enter $1$ and if false then enter $0$

Compute the absolute value of:
$(-5)^4 \times (-5)^6 \div (-5)^{9}$

Compute:
$(9^0 \times 4^{1})$

The value of x,if $\log_{10} x = -2$ is $k$ .
then $100x$ =?

Show that $\displaystyle \left ( 2\sqrt{2} \right )^{-2/3}= \frac{1}{2}.$ can be written as $\displaystyle \log _{2\sqrt{2}} 2= -\frac{2}{3}$

$\displaystyle \log _{3}\frac{1}{243}= -5.$

Show that $\displaystyle 5^{0}= 1$ can be written as $\displaystyle \log _{25}1= 0$

Express $\displaystyle \log _{100}0.1= -\frac{1}{2}$ in exponential form:

$\displaystyle 4^{3/2}= 8$

If $3^{-2}$ is $\cfrac{1}{m}$ , then the value of $m$ is

If $(-4)^{-2}$ is $\cfrac{1}{m}$ , then the value of $m$ is

$(3^0 + 4^{-1}) \times 2^2$

Evaluate $\displaystyle \left ( \frac{5}{8} \right )^{-7} \times \left ( \frac{8}{5} \right )^{-4}$ is $\frac { 512 }{m }$
Value of $m$ is

Evaluate $(5^{-1} \times 2^{-1}) \times 6^{-1}$ is $\frac{1}{m}$
Value of $m$ is

Evaluate $\displaystyle \frac{8^{-1} \times 5^3}{2^{-4}}$

Find the value of $m$ for which $5^m \div 5^{-3} = 5^5$

Evaluate: $-\displaystyle \left \{ \left ( \frac{1}{3} \right )^{-1} - \left ( \frac{1}{4} \right )^{-1} \right \}^{-1}$

Find $x$ , if $(4^{-1} + 8 ^{-1} ) \times (3^{-1} - 9^{-1}) \div \displaystyle \frac{1}{12} = 5^x$

By what number should $\displaystyle \left ( \frac{-2}{3} \right )^{-3}$ be divided so that the quotient may be $\displaystyle \left ( \frac{4}{27} \right )^{-2}$

Simplify: $\left(\cfrac{3}{2} \right)^{0} \times \left(\cfrac{4}{5}\right)^{-2}$

Simplify : $\displaystyle \frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} (t \neq 0)$

Simplify: $3^0 +2^{-2}$

Simplify $\displaystyle \left [ \left ( \frac{2}{3} \right )^{-1} \times \left ( \frac{3}{4} \right )^{-1} \right ]^{-1}$

Find the value $\displaystyle 27^{\cfrac{1}{3} } \times 16^{\cfrac{-1}{4} }$

By what number should $\displaystyle \left ( \frac{3}{4} \right )^{-3}$ be divided so that the quotient becomes $128$ ?

Find the value $(512)^{\frac{-2}{9}}$

For what negative values of $x$ , will $x^{18}$ be equal to $x^{20}$ ?

For what positive values of $x$ , will $x^{18}$ be greater than $x^{20}$ ?

For what negative values of $x$ , will $x^{20}$ be greater than $x^{18}$ ?

If $\displaystyle \log_{3}a=4$ , then find the value of $a$ .

Find the value of x if $x^3 = \displaystyle \left ( \frac{6}{5} \right )^{-3} \times \left ( \frac{6}{5} \right )^6.$

${ 3 }^{ 3 }\times { 3 }^{ 6 }\times { 3 }^{ 7 }=$ ?

Suppose $m$ and $n$ are distinct integers. Can $\cfrac { { 3 }^{ m }\times { 2 }^{ n } }{ { 2 }^{ m }\times { 3 }^{ n } }$ be an integer? Give reasons.

${ 2 }^{ 5 }\times { 5 }^{ 2 }\times { 2 }^{ 3 }\times 5=$ ?

${ 2 }^{ 5 }\times { 2 }^{ 6 }=2^?$

Simplify:
${ 3 }^{ 1 }\times { 3 }^{ 2 }\times { 3 }^{ 3 }\times { 3 }^{ 4 }\times { 3 }^{ 5 }\times { 3 }^{ 6 }$

Simplify:
${2}^{2}\times {3}^{3}\times {2}^{4}\times {3}^{5}\times {3}^{6}$

Simplify and give reasons:
${ \left[ \left( { 3 }^{ 2 }-{ 2 }^{ 2 } \right) \div \cfrac { 1 }{ 5 } \right] }^{ 2 }$

Rakesh solved some problems of exponents in the following way. Do you agree with the solutions? If not why? Justify your argument.
${ x }^{ -3 }\times { x }^{ -2 }={ x }^{ -6 }$

If $\log_{2}x = 3$ , then $x =$ ______

Find the value of ' $x$ ' such that
$\cfrac{1}{49}\times {7}^{2x}={7}^{8}$

Using logarithmic table find the value of the following.
(i) $\log { 23.17 }$
(ii) $\log { 9.321 }$
(iii) $\log { 329.5 }$
(iv) $\log { 0.001364 }$
(v) $\log { 0.9876 }$
(vi) $\log { 6576 }$

Find, by inspection, the characteristics of the logarithms of $21735, 23.8, 350, .035, .2, .87, .875$ .

Given $\log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980$ , find the value of $\log \sqrt [3]{12}$ .

Given $\log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980$ , find the value of $\log 14.4$ .

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log 84$ .

Given $\log 2 = .3010300, \log 3 = .4771213, \log 7 = .8450980$ , find the value of $\log 4\dfrac{3}{2}$ .

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log {^{.}128}.$

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log 64$ .

Give the position of the first significant figure in the numbers whose logarithms are
$\bar{2} {^{.}7781513}, {^{.}6910815}, \bar{5} {^{.}4871384}.$

How many digits are there in the integral part of the numbers whose logarithms are respectively
$4{^{.}30103}, 1{^{.}4771213}, 3{^{.}69897}, {^{.}56515}?$

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log {^{.}0125}.$

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log \sqrt[4]{^{.}0105}.$

Find the product of $37.203, 3.7203, .037203, 372030$ , having given that
$\log 37.203 = 1.5705780$ , and $\log 1915631 = 6.2823120$ .

Evaluate : $2^5\times 2^8\div 2^6$

Given $\log2 = {^{.}3010300}, \log3 = {^{.}4771213}, \log7 = {^{.}8450980},$ find the value of
$\log \sqrt{\dfrac{35}{27}}.$

Solve the following equations.
$\displaystyle log_2 \, (4^x \, + \, 4) \, - \, log_2 \, (2{x \, + \, 1} \, - \, 3) \, = \, 0.$

Write value of $\sqrt [ 3 ]{ 2 } \times \sqrt [ 4 ]{ 2 } \times \sqrt [ 12 ]{ 32 }$

Solve the following equation:
$\displaystyle\, 5^{x - 1} = 10^x\cdot 2^{-x}\cdot 5^{x + 1}$

Rewrite the following equation in the logarithm from :
$5^0 \, = \, 1$

Rewrite the following equation in the logarithm from :
$(2\sqrt{2})^{-2/3} \, = \, \dfrac{1}{2}.$

Rewrite the following equation in the exponential form.
$log_{5\sqrt{5}} \, 5 \, = \, \dfrac{2}{3}$

Rewrite the following equation in the exponential form.
$log_{100} \, 0.1 \, = \, -\dfrac{1}{2}$

Solve $x^{\sqrt{x}} \, = \, (\sqrt {x})^x$

Find the value of ${(23.17)^{{1 \over {5.76}}}}$ using log table.

Rewrite the following equation in the exponential form.
$\log_3 \, \frac{1}{243} \, = \, -5.$

Evaluate : ${2^{{{\log }_3}5}} - {5^{{{\log }_3}2}}$

${ \left[ { \left( \dfrac { 15 }{ 12 } \right) }^{ 3 } \right] }^{ 4 }$

If $x=243$ , then find the value of ${x^{\frac{1}{5}}} \times {x^{ - \frac{1}{5}}}$

Solve: $\log (3x + 2) + \log (3x - 2) = \log5$

Convert the following to logarithmic form:
$5^{2} =25$

If $X^\cfrac{a}{b} =1,$ then find the value of $'a'$ .

Find the value of x for which
${\left( {{3 \over 4}} \right)^6} \times {\left( {{{16} \over 9}} \right)^5} = {\left( {{4 \over 3}} \right)^{x + 2}}$

Simplify:
$\log_{10}5 + 2\log_{10}4$