Logarithm And Antilogarithm - Class 11 Commerce Applied Mathematics - Extra Questions

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 $$2\sqrt2$$
$$\displaystyle \log _{2\sqrt{2}} (\frac{1}{2})= -\frac{2}{3}$$



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

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

  $$\displaystyle (-4)^{-2}$$
$$ = \frac{1}{(-4)^2}$$
$$ = \frac{1}{(-1)^2 \times (4)^2}$$
$$ = \frac{1}{1 \times 16}$$
$$ = \frac{1}{16}$$


Find the value of: $$(3^0 + 4^{-1}) \times 2^2$$

    $$\displaystyle (3^0 + 4^{-1}) \times 2^2$$
$$ = \left ( 1 + \frac{1}{4} \right ) \times 4$$
$$ = \frac{4 + 1}{4} \times 4$$
$$= \displaystyle \frac{5}{4} \times 4$$
$$ = 5$$


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 $$x>0$$, 
If   $$x=\cfrac{p}{q}$$ 
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 $$x^{18}$$, 
If  $$x=-\cfrac{p}{q}$$, 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:
$${ \left[ \left( { 3 }^{ 2 }-{ 2 }^{ 2 } \right) \div \cfrac { 1 }{ 5 }  \right]  }^{ 2 }$$


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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 $${8}^{x}={2}^{3x}$$
so $${8}^{3a}={2}^{9a}$$
$${8}^{3a}\times{2}^{5}\times{2}^{2a}={2}^{9a}\times {2}^{2a+5}={2}^{11a+5}$$
$${4}={2}^{2}$$
$$4\times{2}^{11a}\times{2}^{-2a}={2}^{2}\times{2}^{9a}={2}^{9a+2}$$
$$\dfrac{{2}^{11a+5}}{{2}^{9a+2}}={2}^{2a+3}$$


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

$$=3.968(\sqrt{3.968})$$

$$=3.968(\sqrt{3.968})$$

$$=3.968\times 1.99$$

$$=7.9041$$

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

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


 Using laws of exponents, simplify and write the answer in exponential form: $$\left( { 3 }^{ 4 } \right) ^{ 3 }$$

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 $$a=10,b=x,x=y$$
So $$y=\log _{10}x\implies x=10^y$$


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,
$$3\log 5 - 3\log 3 - \log 5 + 2 \log 6 = 2 - \log n$$
$$2 \log  5 - 3 \log 3 + 2 \log 6 = 2 (1) - \log n$$
$$\log 5^{2} - \log 3^{3} + \log 6^{2} = 2 \log 10 - \log n[Since, 1 = \log 10]$$
$$\log n = - \log 25 + \log 27 - \log 36 + \log 100$$
$$= (\log 100 + \log 27) - \log 5 + \log 36)$$
$$= \log  (100 \times 27) - \log  (25 \times 36)$$
$$= \log (100 \times 27) - \log  (25 \times 36)$$
$$= \log (100 \times 27)/(25 \times 36)$$
$$\log n = \log 3$$
$$n = 3$$


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,
$$ \log_{10}\left ( x +2 \right ) + \log_{10}\left ( x-2 \right ) = \log_{10}3+3 \log_{10}4$$ 
$$\log_{10} [(x + 2) \times (x - 2)] = \log_{10} 3 + \log_{10} (3 \times 4^{3})$$
$$[(x +2) \times (x - 2)] = (3 \times 4^{3})$$
$$[(x +2) \times (x - 2)] = (3 \times 4^{3})$$
$$(x^{2} - 4) = (3 \times 4 \times 4 \times 4)$$
$$(x^{2} - 4) = 192$$
$$x^{2} = 192 + 4$$
$$= 196$$
$$x = \sqrt{196}$$
$$= 14$$


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 $$A$$
$$\implies A=a^x$$.


If $$\log 2 = 0.3010$$ and $$\log 3 = 0.4771$$, then the value of $$\log 3.6$$ 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)


If $$\displaystyle \log_{10} x = 2a$$ and $$\displaystyle \log_{10} y = \tfrac {b}{2}$$, then $$\displaystyle 10^a$$ 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 $$5.$$

752398_616698_ans_d295047b3972490e92ab9f1b79a252ec.png


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.