MCQ Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 6

If the value of

$(\log_{10} 2 )+ 1$ is in the form of

$\log_{10}m$ , then

$m$ is equal to

0%
$11$
0%
$14$
0%
$20$
0%
$17$

Explanation

$\log _{ 10 }{ 2 } +1 = \log _{ 10 }{ 2 } +\log _{ 10 }{ 10 } \\ \\ \\ =\log _{ 10 }{ (2\times 10 } )\\ = \log _{ 10 }{ 20 }$

The value of

$\displaystyle \log_{10}0.001$ is equal to

0%
$-3$
0%
$3$
0%
$-2$
0%
$2$

Explanation

Consider,

$\log _{ 10 }{ 0.001 } =x$

$\implies 0.001 = { 10 }^{ x }$

$\implies \dfrac { 1 }{ 1000 } = 10^x$

$\implies { 10 }^{ -3 }= 10^x$

$\implies x=-3$

$\log (a)^{3} + \log a$ =

$m \log a$ , then value of

$m$ is equal to

0%
$1$
0%
$4$
0%
$2$
0%
$-5$

Explanation

$\textbf{Step 1 : Use the rule of logarithm}$

$\text{According to logarithm power rule,}$

$\because \log_e( X ^Y) = Y . \log_e(X)$

$\text{Therefore, }\log(a)^3 = 3. \log \:a$

$\text{Since,} \log (a)^3 + \log a = m \log a$

$\text{Therefore, }3. \log a + \log a=m \log a$

$4 \log a=m \:\log\:a$

$\text{By comparing both the sides we get,}$

$m=4$

$\textbf{Hence, option B is correct.}$

The logarithm form of

$\displaystyle (81)^\frac{3}{4} = 27$ is

$\log_{81} 27 = \displaystyle \frac{3}{m}$ . Then value of

$m$ is equal to

0%
$2$
0%
$0$
0%
$4$
0%
$3$

Explanation

Given,

$\log_{81} 27 = \displaystyle \frac{3}{m}$ ...(1)

Now,

$\displaystyle (81)^\frac{3}{4} = 27$

Converting in logarithm form, we can write

$\log_{81} 27 = \displaystyle \frac{3}{4}$ . ...(2)

From (1) and (2), we get

$\displaystyle \dfrac{3}{m}=\displaystyle \dfrac{3}{4}$

$\therefore m=4$

$\dfrac{\log 81}{\log 27} = x$ , then the value of

$x$ is equal to

0%
$\dfrac { 4 }{ 3 } \\ \\$
0%
$\dfrac { 3 }{ 4 } \\ \\$
0%
$\dfrac { 1 }{ 3 } \\ \\$
0%
Not solvable

Explanation

$\dfrac{\log 81}{\log 27} = x$

$\Rightarrow x = \dfrac{\log 3^4}{\log 3^3} = \dfrac{4 \log 3}{3 \log 3} = \dfrac{4}{3}$

The logarithm form of

$10^{-3} = 0.001$ is

$\log_{10} 0.001 = -m$ , then value of

$m$ is

0%
$-1$
0%
$-2$
0%
$3$
0%
$-4$

Explanation

Given,

${ 10 }^{ -3 }= 0.001$

$\implies \log _{ 10 }{ 0.001 } = -3$

$\therefore m = 3$ .

$\log_{10}(x - 10) = 1$ , then value of

$x$ is

0%
$10$
0%
$13$
0%
$20$
0%
$26$

Explanation

Since,

$\log _{ 10 }{ (x-10) } =1$

$\implies (x -10)=10^1$

$\implies x = 20$ .

$\displaystyle 64^{a}=\frac{1}{256^{b}}$ , then

$3a + 4b$ equals:

0%
$2$
0%
$4$
0%
$8$
0%
$0$

Explanation

$64^{a}=\dfrac{1}{256^{b}}$

$\Rightarrow (2^{6})^{a}=\dfrac{1}{(2^{8})^{b}}$

$\Rightarrow 2^{6a}\times 2^{8b}=1$

$\Rightarrow 2^{6a+8b}=2^{0}$

$\Rightarrow 6a+8b=0$

$\Rightarrow 3a+4b=0$

$x = 2$ and

$y = 3$ , then find the value of

$\left[ \displaystyle\frac { 1 }{ x^{ x } } +\displaystyle\frac { 1 }{ y^{ y } } \right]$ .

0%
$\displaystyle\frac { -31 }{108 }$
0%
$\displaystyle\frac { 31 }{108 }$
0%
$\displaystyle\frac { 125 }{171}$
0%
$\displaystyle\frac { 153 }{222}$

Explanation

$\cfrac { 1 }{ x^{ x } } +\cfrac { 1 }{ y^{ y } } =\cfrac { 1 }{ 2^{ 2 } } +\cfrac { 1 }{ 3^{ 3 } }$

$=\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 27 } \\ =\cfrac { 27+4 }{ 108 }$

$=\cfrac { 31 }{ 108 }$

$\log_4m\,=\,1.5$ , then the value of

$m$ is equal to

0%
$m=8$
0%
$m=4$
0%
$m=16$
0%
$m=64$

Explanation

Since,

$\log _{ 4 } m\, =\, 1.5$

$\Rightarrow m={ 4 }^{ 1.5 }=2^{2\times 1.5}=2^3=8$

The value of

$[ 5 (8^{\tfrac 13} + 27^{\tfrac 13} )^3 ]^{\tfrac 14}$ is

0%
$5^4$
0%
$5^{\tfrac 14}$
0%
$5$
0%
None of these

Explanation

$\left [ 5 \left ( 2^{3\times \tfrac {1}{3}}+ 3^{3\times \tfrac {1}{3}} \right )^3 \right ]^{\tfrac 14}$

$=[ 5 (2 + 3)^3 ] ^{\tfrac 14}$

$= (5 \times 5^3)^{\tfrac 14} = 5^{4\times \tfrac {1}{4}} = 5$

$\displaystyle \sqrt{3^{n}}=81$ . Then n is equal to

0%
2
0%
4
0%
6
0%
8

Explanation

$\Rightarrow \sqrt{3^{n}}=81$

$\Rightarrow 3^{n/2}=3^{4}$
Base are same so

$\Rightarrow \frac{n}{2}=4\Rightarrow n=8$

The value of

$512^\frac {-2}{9}$ is

0%
$\displaystyle \frac {1}{2}$
0%
$2$
0%
$4$
0%
$\displaystyle \frac {1}{4}$

Explanation

$\begin{aligned}{}{\left( {512} \right)^{ - {\textstyle{2 \over 9}}}} &= {\left( {{2^9}} \right)^{ - {\textstyle{2 \over 9}}}}\\&={2^{9 \times \frac{{ - 2}}{9}}} \quad\quad\quad\because[(a^b)^c=a^{bc}]\\&= {2^{ - 2}}\\ &= \frac{1}{4}\end{aligned}$

Hence, option

$D$ is correct.

The value of

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

0%
$109\, \displaystyle \frac {2}{9}$
0%
$109\, \displaystyle \frac {9}{2}$
0%
$109$
0%
None of these

Explanation

Given expression is

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

Using the law of exponent,

$a^{-m}=\dfrac{1}{a^m}$ , we can write it as:

$=1 - (-27) - \left(-\cfrac {1}{3}\right)$

$+ 81 - \cfrac 19$

$= 1 + 27 + \cfrac {1}{3} + 81 - \cfrac {1}{9}$

$=109 + \cfrac {2}{9}\,$

$=\, 109 \displaystyle \frac {2}{9}$

Hence, option A is correct.

SImplify:

$(256)^{0.16} \times (256)^{0.09}$

0%
$4$
0%
$16$
0%
$64$
0%
$256.25$

Explanation

$(256)^{0.16} \times (256)^{0.09}$

$=(256)^{(0.16 + 0.09)}$

$= (256)^{0.25} = (256)^{\tfrac {25}{100}} = (256)^{\tfrac 14}$

$= (4^4)^{\tfrac 14} = 4^1 = 4$

The value of

$4(7^1.\, 7^{-1}.\, 7^{-1}.7^0)$ is

0%
$3\displaystyle \frac {7}{3}$
0%
$\displaystyle \frac {6}{7}$
0%
$3\displaystyle \frac {3}{7}$
0%
None of these

The value of

$[10^{150}\, \div\, 10^{146} ]$ is

0%
$1000$
0%
$10000$
0%
$100000$
0%
$10^5$

Explanation

$(10)^{150} \div (10)^{146} = \cfrac {(10)^{150}}{(10)^{146}}$

$= 10^{(150 - 146)} = 10^4 = 10000$

$\log_4$ 1 is equal to

0%
$1$
0%
$0$
0%
$\infty$
0%
none of these

Explanation

$log_a 1 = m$

$\Rightarrow a^m = 1 a^0 \Rightarrow m = 0$

Value of

$\displaystyle \log _{4}18$ is:

0%
an irrational number
0%
a rational number
0%
natural number
0%
whole number

Explanation

$\log_{4}{18}=\log_{2^{2}}{(2.3^{2})}=\dfrac{1}{2}\log_{2}(2.3^{2})$

$=\dfrac{1}{2}\left [ \log_{2}{2}+\log_{2}{3^{2}} \right ]=\dfrac{1}{2}\left [ 1+2.\log_{2}{3} \right ]$

$\log_{2}{3}$ is an irrational number.

Hence,

$\dfrac{1}{2}\left [ 1+2.\log_{2}{3} \right ]$ is also an irrational number.

The value of

$(6^{-1} -7^{-1})^{-1} (5^{-1} -4^{-1})^{-1}$ is equal to

0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{1}{4}$
0%
None of these

Explanation

Given expression is

$\left({ 6 }^{ -1 }-7^{ -1 }\right)^{ -1 }\left(5^{ -1 }-4^{ -1 }\right)^{ -1 }$ .

From laws of exponents, we know that

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

Hence, the given expression can be simplified as:

$\left(\cfrac { 1 }{ 6 } -\cfrac { 1 }{ 7 } \right)^{ -1 }\left(\cfrac { 1 }{ 5 } -\cfrac { 1 }{ 4 } \right)^{ -1 }$

Solving the fractions, we get:

$\cfrac { 1 }{ 6 } -\cfrac { 1 }{ 7 }=\cfrac{1}{42}$

and

$\cfrac { 1 }{ 5 } -\cfrac { 1 }{ 4 }=\cfrac{-1}{20}$

Hence, the expression simplifies to:

$\left(\cfrac { 1 }{ 42 } \right)^{ -1 }\left(-\cfrac { 1 }{ 20 } \right)^{ -1 }$

Again applying the law of exponent,

$a^{-m}=\left(\dfrac{1}{a}\right)^m$ , we get:

$(42)^1(-20)^1$

$\Rightarrow 42\times (-20)=-840$

Hence,

$\left({ 6 }^{ -1 }-7^{ -1 }\right)^{ -1 }\left(5^{ -1 }-4^{ -1 }\right)^{ -1 }=-840$ .

Value of the expression

$\displaystyle \log _{2}\sqrt[5]{2.\sqrt[3]{2\sqrt{2}}}$ is

0%
$0.1$
0%
$0.2$
0%
$0.3$
0%
does not exist

Explanation

$\log_{2}\left ( \sqrt[5]{2.\sqrt[3]{2\sqrt{2}}} \right )=$

$=\log_{2}\left ( \sqrt[5]{2.({2^{3/2}})^{1/3}} \right )$

$=\log_{2}\left ( \sqrt[5]{2.2^{1/2}} \right )$

$=\log_{2}\left ( 2^{3/2} \right )^{1/5}=\dfrac{1}{5}\, \log_{2}.{2^{3/2}}$

$=\dfrac{3}{10}.\log_{2}{2}=\dfrac{3}{10}=0.3$

Simplify :

$(6^{-1} - 8^{-1})^{-1} + (2^{-1} - 3^{-1})^{-1}$ is-

0%
$20$
0%
$30$
0%
$10$
0%
$40$

Explanation

Law of exponents,

Given -

$(6^{-1}-8^{-1})^{-1}+(2^{-1}-3^{-1})^{-1}$

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

$=\left(\dfrac{8-6}{6\times 8}\right)^{-1}+\left(\dfrac{3-2}{2\times 3}\right)^{-1}$

$=\left(\dfrac{2}{36\times 8}\right)^{-1}+\left(\dfrac{1}{2\times 3}\right)^{-1}$

$=\left(\dfrac{1}{24}\right)^{-1}+\left(\dfrac{1}{6}\right)^{-1}$

$=24+6$

$=30$

Hence, option B is correct.

$\left (\dfrac {a}{b}\right )^{x-1}=\left (\dfrac {a}{b}\right )^{x-3}$ then the value of

$x$ is

0%
-1
0%
1
0%
2
0%
3

Explanation

Given,

$\left (\dfrac {a}{b}\right )^{x-1}=\left (\dfrac {a}{b}\right )^{-(x-3)}$

as the terms are same we can equate the powers/exponents of the terms

$\Rightarrow x-1=-(x-3)$

$\Rightarrow x-1=-x+3$

$\Rightarrow 2x=4$

$\Rightarrow x=2$

$x=\log _{ a }{ bc } ,y=\log _{ b }{ ca } ,z=\log _{ c }{ ab }$ , then the value of

$\dfrac { 1 }{ 1+x } +\dfrac { 1 }{ 1+y } +\dfrac { 1 }{ 1+z }$ will be

0%
$x+y+z$
0%
$1$
0%
$ab+bc+ca$
0%
$abc$

Explanation

Now,

$1+x=\log _{ a }{ a } +\log _{ a }{ bc } =\log _{ a }{ abc }$

$\Rightarrow \dfrac { 1 }{ 1+x } =\log _{ abc }{ a }$
Similarly,

$\dfrac { 1 }{ 1+y } =\log _{ abc }{ b }$ and

$\dfrac { 1 }{ 1+z } =\log _{ abc }{ c }$

$\therefore \dfrac { 1 }{ 1+x } +\dfrac { 1 }{ 1+y } +\dfrac { 1 }{ 1+z }$

$=\log _{ abc }{ a } +\log _{ abc }{ b } +\log _{ abc }{ c }$

$=\log _{ abc }{ abc } =1$

$\displaystyle { log }_{ 5 }{ log }_{ 5 }{ log }_{ 2 }x=0$ , then the value of

$x$ is

0%
$32$
0%
$125$
0%
$625$
0%
$25$

Explanation

Given,

$\displaystyle { log }_{ 5 }{ log }_{ 5 }{ log }_{ 2 }x=0$

$\displaystyle \Rightarrow \quad { log }_{ 5 }{ log }_{ 2 }x=50$

$\displaystyle \Rightarrow \quad { log }_{ 5 }{ log }_{ 2 }x=1\Rightarrow { log }_{ 2 }x=5$

$\displaystyle \Rightarrow x={ 2 }^{ 5 }=32$

The value of

$\displaystyle \log_{2}\left [ \log_{2}\left \{ \log_{3}\left ( \log_{3}27^{3} \right ) \right \} \right ]$ is equal to

0%
2
0%
3
0%
0
0%
1

Explanation

Required value of

$\displaystyle \log _{2}\left [ \log _{2}\left \{ \log _{3}\left ( \log _{3}27^{3} \right ) \right \} \right ]$

$=$

$\displaystyle \log _{2}\left [ \log _{2}\left \{ \log _{3}\left ( \log _{3}3^{9} \right ) \right \} \right ]$

$=$

$\displaystyle \log _{2}\left [ \log _{2}\left \{ \log _{3}\left (9\cdot \log _{3}3 \right ) \right \} \right ]$

$=$

$\displaystyle \log _{2}\left [ \log _{2} \left \{\log _{3}\cdot 9 \right \}\right ]$ ....

$\displaystyle \left [ \because \log _{3}3=1 \right ]$

$=$

$\displaystyle \log _{2} \left [ \log _{2}\left \{ \log _{3}3^{2} \right \} \right ]$

$=$

$\displaystyle \log _{2} \left [ \log _{2} 2\cdot \log _{3} 3\right ]$

$=$

$\displaystyle \log _{2}\left [ \log _{2}2\right ]$ ....

$\displaystyle \left [ \because \log _{2}2=1 \right ]$

$=$

$\displaystyle \log _{2}$

$1 = 0$

If there are

$n$ zeros after the decimal point, then the characteristic of that number will be

0%
$n+1$
0%
$-n+1$
0%
$-(n+1)$
0%
$n-1$

Explanation

For number less than

$1,$ if there are

$n$ zeroes after the decimal point, then the characteristics of that number will be

$-(n+1).$

and For number greater than

$1,$ if there are

$n$ number of zeroes are on the left sides of the digits then characteristics will be

$(n+1).$

Hence, C is the correct option.

The characteristic of a number having

$m$

$(m>1)$ digits is given by,

0%
$m-1$
0%
$m+1$
0%
$m$
0%
None of the above

Explanation

If a number is

$N>0$ , then

$\log_{10}N$ will have two parts, the integral part is known the characteristic and the decimal part is known as mantissa.

$2$ digit belongs from

$[10,100]$ where

$\log_{10}10=1$ and

$\log_{10}100=2$

Similarly,

$m$ digit number belongs from

$[10^{m-1},10^m]$ , where

$\log_{10}10^{m-1}=m-1$ and

$\log_{10}10^m=m$ .

Thus any number between

$[10^{m-1},10^m]$ will have

$m-1$ as the integral part.

Thus the characteristic of a number having

$m$ digits is given by

$m-1$ .

Let

$x = (0.15)^{20}$ . Find the characteristic in the logarithm of

$x$ to the base

$10$ .

0%
$17$
0%
$21$
0%
$-21$
0%
$-17$

Explanation

Given

$x=(0.15)^{20}$

By applying

$\log$ on both sides , we get

$\log _{ 10 }{ x } =20\log _{ 10 }{ (0.15) } =-16.478$

$\Rightarrow \log _{ 10 }{ x } =-17+0.5218$

The integral part of

$\log _{ 10 }{ x }$ is called characteristic

Therefore the characteristic of given number is

$-17$

So option

$D$ is correct

The value of

$\log_{10} 0.0006024$ is equal to

0%
$\overline {3}.7979$
0%
$\overline {1}.9779$
0%
$\overline {4}.7799$
0%
$0.7279$

Explanation

$\log_{10}0.0006024$

Characteristics

$=-4$

For mantissa ,read from the table

$6024.$

Mantissa

$=7799$

Thus,

$\log_{10}0.0006024=$ characteristics of

$0.0006024+$ mantissa of

$0.0006024$

$=-4+0.7799$

$=\bar4.7799$

Hence, C is the correct option.

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 6 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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