MCQ Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 1

Find the characteristic of

$\log 27.93$

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

Explanation

From log table,

$\log27.93=1.4460$

Here, Characteristics

$=1$ and Mantisa

$=0.4460$

Hence, B is the correct option.

Find the characteristic of

$\log 7.93$

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

Explanation

From Logarithmic table,

$\log7.93=0.89927$

Here, Characteristics

$=0$

Hence, A is the correct option.

Find the characteristic of

$\log 277.9301$

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

Explanation

From logarithmic table,

$\log 277.9301=2.4439$

Here, Characteristics

$=2$

Hence, C is the correct option.

The value of

$10^{\log_{10}{10}^{7}}$ is:

0%
$7$
0%
$10^7$
0%
$10$
0%
$\log_{10}7$

Explanation

Consider,

$10^{\log_{10}{10}^{7}}$

$= 10^{{{7\log}_{10}}{10}}$

$= 10^{7\times 1}$

$= 10^{7}$ .

Solve the following using Product Law of Exponents.

$a\times { a }^{ 2 }\times { a }^{ \tfrac { 1 }{ 2 } }$

0%
${ a }^{ 7 }$
0%
${ a }^{ \tfrac { 2 }{ 7 } }$
0%
${ a }^{ 3 }$
0%
${ a }^{ \tfrac { 7 }{ 2 } }$

Explanation

By product law of exponents

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

$={ a }^{ \tfrac { 7 }{ 2 } }$
Therefore option

$D$ is the correct answer.

$\log_{10} x = a$ and

$\log_{10} y = b$ , then

$10^{a-1}$ in terms of

$x$ will be

0%
$\displaystyle \frac{x}{7}$
0%
$\displaystyle \frac{x}{8}$
0%
$\displaystyle \frac{x}{9}$
0%
$\displaystyle \frac{x}{10}$

Explanation

Given,

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

$\implies 10^a=x$

$\implies 10^{a-1}=\dfrac{x}{10}$

Hence, option D is correct.

Simplify:

$\left((9)^{0} + (11)^{0} + (13)^{0}\right) \div (23)^{0}$

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

Explanation

$\left((9)^{0} + (11)^{0} + (13)^{0}\right) \div (23)^{0}$

$= (1 + 1 + 1) \div 1$

$= 3\div 1$

$= 3$

So, option

$B$ is correct.

The value of

$x$ satisfying

$\log _{ 243 }{ x } =0.8$

0%
$81$
0%
$1.8$
0%
$2.43$
0%
$27$

Explanation

$\log_{243}x=0.8$

$\Rightarrow x=243^{0.8}$

$\Rightarrow x=81$

Hence, A is the correct option.

$7^{10}= 7 \times 7^n$ , what is the value of

$n$ ?

0%
$10$
0%
$9$
0%
$7$
0%
$5$
0%
$3$

Explanation

Given,

${ 7 }^{ 10 }={ 7 }^{ 1 }\times { 7 }^{ n }$

$\Rightarrow 7^{10}={ 7 }^{ 1+n }$ .

$\Rightarrow 10 = 1+n$ (as bases are equal, their powers must be equal)

$\Rightarrow n = 9$

The value of

$x$ satisfying the logarithm

$\log_{243} x = 0.8$ is equal to

0%
$81$
0%
$1.8$
0%
$2.43$
0%
$27$

Explanation

Given,

$\log_{243} x = 0.8$

$\Rightarrow 243^{0.8} = x$

$\Rightarrow (3^5)^{4/5} = x$

$\Rightarrow 3^4 = x$

$\Rightarrow x = 81$

Given that

$4^{n+1} = 256$ , find the value of

$n$ .

0%
$2$
0%
$3$
0%
$5$
0%
$63$

Explanation

$4^{n+1}=256$

Taking log on both sides to the base 2

$\Rightarrow\: \log_{2}{4^{n+1}}=\log_{2}{256}$

$\Rightarrow\:(n+1)\log_{2}{4}=\log_{2}{2^{8}}$

$\Rightarrow\:(n+1).\log_{2}{2^2}=8$

$\Rightarrow\:(n+1).2=8$

$\Rightarrow\:(n+1)=4$

$\Rightarrow\:n=3$

The value of

$7^{\frac{1}{2}}. 8^{\frac{1}{2}}$ is :

0%
$28^{\frac{1}{2}}$
0%
$56^{\frac{1}{2}}$
0%
$14^{\frac{1}{2}}$
0%
$42^{\frac{1}{2}}$

Explanation

$\Rightarrow$

$(7)^{\tfrac{1}{2}}.(8)^{\frac{1}{2}}$

$\Rightarrow$

$(7\times 8)^{\frac{1}{2}}$

$[\,\because\,m^x.n^x=(m\times n)^x]$

$\Rightarrow$

$56^{\frac{1}{2}}$

Find the product.

$(a^2) (2a^{22}) (4a^{26})$

0%
$8a^{40}$
0%
$8a^{50}$
0%
$8a^{30}$
0%
$8^{20a}$

Explanation

$(a^2) (2a^{22}) (4a^{26})$

$=(a^2)\times (2a^{22})\times (4a^{26})$

$=8a^{2+22+26}$

$=8a^{50}$

The value of

$\cfrac{3^0+7^0}{5^0}$ is:

0%
$2$
0%
$0$
0%
$\dfrac{9}{5}$
0%
$\dfrac{1}{5}$

Explanation

$\Rightarrow$

$\dfrac{3^0+7^0}{5^0}$

$\Rightarrow$

$\dfrac{1+1}{1}$

$[\,\because\,x^0=1]$

$\Rightarrow$

$\dfrac{2}{1}$

$\Rightarrow$

$2$

$\therefore$

$\dfrac{3^0+7^0}{5^0}=2$

Simplified value of

$(25)^{\frac{1}{3}} \times (5)^{\frac{1}{3}}$ is :

0%
$25$
0%
$3$
0%
$1$
0%
$5$

Explanation

$(25)^{\tfrac{1}{3}}\times (5)^{\tfrac{1}{3}}$

$\Rightarrow$

$(25\times 5)^\tfrac{1}{3}$ ...

$[\because\,m^x\times n^x=(m\times n)^x]$

$\Rightarrow$

$(125)^\tfrac{1}{3}$

$\Rightarrow$

$\sqrt[3]{125}=5$

$\therefore$

$(25)^{\tfrac{1}{3}}\times (5)^{\tfrac{1}{3}}=5$

State true or false:

$(\cfrac{2}{3})^4 \times (\cfrac{27}{8})^{-2}= \cfrac{4}{9}$

0%
True
0%
False

Explanation

$(\cfrac{2}{3})^4 \times (\cfrac{27}{8})^{-2}$

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

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

$=(\cfrac{2}{3})^4 \times (\cfrac{3}{2})^{(-6)}$

$=(\cfrac{2}{3})^4 \times (\cfrac{2}{3})^{6}$

$=(\cfrac{2}{3})^{(4+6)}$

$=(\cfrac{2}{3})^{10}$

The given statement is false.

State true or false:

$\displaystyle x^y = z$ , then

$\displaystyle y = log_z x$ .

0%
True
0%
False

Explanation

Since,

$x^y=z$

$\implies \log$ of

$z$ to the base

$x$ is

$y$

$\implies y=log_xz$
Hence, given statement is false.

$\displaystyle \log_3 x = 0$ , then value of

$x$ is equal to

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

Explanation

Given,

$\log_3x = 0$

$\implies x=3^0$ (Since we know that

$\log_ab=t\Rightarrow b=a^t$ )

$\therefore x=1$ .

The value of

$\displaystyle \log_5 1$ is

0%
$0$
0%
$1$
0%
$3$
0%
$5$

Explanation

Consider,

$\log _{ 5 }{ 1 } =x$

$\implies 1 = { 5 }^{ x }$

$\implies 5^0 = 5^x$

$\implies x = 0$ .

The exponential form of

$\log_8 0.125 = -1$ is

$8 ^{-m} = 0.125$ . Then value of

$m$ is

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

Explanation

$\log _{ 8 }{ 0.125 } = -1\\ \implies 0.125 = { 8 }^{ -1 }$

Also, by given we have

$8^{-m} = 0.125$

Therefore

$m\ =\ 1$

The value of

$\log_2 \displaystyle \frac{1}{8}$ is equal to

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

Explanation

Given,

$\log _{ 2 }{ \frac { 1 }{ 8 } } =x$

Convert in exponential form,

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

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

$\therefore x = -3$

The exponential form of

$\log_{10}1 = 0$ is

$10^{m} = 1$ , then the value of

$m$ is

0%
$2$
0%
$0$
0%
$1$
0%
$6$

Explanation

$\log _{ 10 }{ 1 }= 0\\ 1 = { 10 }^{ 0 }$

The logarithm of

$27$ to the base

$9$ is

$\displaystyle \frac{3}{m}$ . Then the value of

$m$ is equal to

0%
$0$
0%
$2$
0%
$3$
0%
$6$

Explanation

Since,

$\log _9 {27}= \displaystyle \frac {3}{m}$ ...(1)

Let

$\log _{ 9 }{ 27 } = x$

Converting in exponential form,

$27 = { 9 }^{ x }$

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

$x=\displaystyle \frac { 3 }{ 2 }$ ...(2)

Comparing equations (1) and (2), we get

$m=2$

If exponential form of

$\log_{10} 0.01 = -2$ is

$10^{m} = 0.01$ , then value of

$m$ is equal to

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

Explanation

$\log _{ 10 }{ 0.01 } = -2 \Rightarrow 0.01 = { 10 }^{ -2 }$

$\therefore m=-2$

The value of

$\log_{10}0.01$ is equal to

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

Explanation

$\log _{ 10 }{ 0.01= } \log _{ 10 }{ { 10 }^{ -2 } }$

$\implies \log _{ 10 }{ 0.01= } -2$ .

$\log_{x} 2 = -1$ then value of

$2x$ is equal to

0%
$2$
0%
$1$
0%
$0$
0%
$5$

Explanation

Given,

$\log _{ x }{ 2 } = -1$

Converting in exponential form,

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

$2 = \displaystyle \dfrac { 1 }{ x }$

$\therefore x=\displaystyle \dfrac {1}{2}$

$2 x=1$

Value of

$\sqrt [4]{(81)^{-2}}$ is

0%
$\dfrac {1}{9}$
0%
$\dfrac {1}{3}$
0%
$9$
0%
$\dfrac {1}{81}$

Explanation

As given

$\sqrt[4]{{(81)}^{-2}}$

$=\sqrt[4]{\dfrac{1}{\left ( 81 \right )^{2}}}$

$=\sqrt[4]{\dfrac{1}{\left ( \left (9 \right )^{2} \right )^{2}}}$

$=\dfrac{\left ( 1 \right )^{\frac{1}{4}}}{\left ( \left ( 9 \right )^{4} \right )^{\frac{1}{4}}}$

$=\dfrac{1}{9}$

The value of

$\log_5\ 125$ is equal to

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

Explanation

$\text {Consider}, \log _{ 5 }{ 125 } =x\\ \implies 125 = { 5 }^{ x }\\ \implies 5^3 =5^x \\\implies x = 3$

Value of

$\displaystyle\frac{{2}^{100}}{2}$ is-

0%
$1$
0%
${50}^{100}$
0%
${2}^{50}$
0%
${2}^{99}$

Explanation

$\Rightarrow \dfrac{{2}^{n}}{2}={2}^{n-1}$

$\Rightarrow \dfrac{{2}^{100}}{2}={2}^{100-1}={2}^{99}$
Hence option 'D' is correct.

Express the following in logarithmic form

$\,\colon$

$81\,=\,3^{4}$

0%
$\log_381\,=\,4$
0%
$\log_981\,=\,2$
0%
$2\log_39\,=\,4$
0%
$4\log_93\,=\,2$

Explanation

$y={ a }^{ x }\Rightarrow \log _{ a }{ y } =x\\ \therefore 81={ 3 }^{ 4 }\Rightarrow \log _{ 3 }{ 81 } =4$

A is true.

$81=3^{4}=9^{2}$

$\Rightarrow \log_{9}81=\log_{9}9^{2}=2\log_{9} 9=2$

B is true.

$81=3^{4}=9^{2}$

$\Rightarrow \log_{3} 3^{4}=\log_{3}9^{2}=2\log_{3} 9$

C is true.

$81=3^{4}=9^{2}$

$\Rightarrow \log_{9}3^{4}=\log_{9}9^{2}=2\log_{9} 9=2$

$\Rightarrow 4\log_{9} 3=2$

D is true.

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 1 - 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 1 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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