MCQ Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 5

$\log_{10}(x+5) = 1$ , then value of

$x$ is equal to

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

$\log_{10}(x+5) = 1$

$\mbox{In exponent form},$

$(x+5) = { 10 }^{ 1 }$

$\therefore x = 10 - 5 = 5$

Given log 6 and log 8, then the only logarithm that cannot be obtained without using the table is

0%
$\log 64$
0%
$\log 21$
0%
$\log \frac{8}{3}$
0%
$\log 9$

Explanation

Let $\log { 6 } =x$

$\Rightarrow \log { 2 } +\log { 3 } =x$ ....... $(i)$

Let $\log { 8 } =y$

$\Rightarrow 3\log { 2 } =y$

$\Rightarrow \log 2=\dfrac{y}{3}$ ...... $(ii)$

$\therefore x=\log { 3 } +\dfrac { y }{ 3 }$ ...... From $(i)$ and $(ii)$

$\Rightarrow \log { 3 } =x-\dfrac { y }{ 3 }$

$\log { 64 } =\log { { 2 }^{ 6 } }=6\log 2=6\dfrac{y}{3}$

$\log { 21 } =\log { 3 } +\log {7}$

Thus, $\log 21$ cannot be found from given values as we need the value of $\log { 7 }$ .

$\log \dfrac{8}{3}=\log 8-\log 3=3\log 2-\log 3$

$\log 9=\log 3^{2}=2\log 3$

Hence, only $\log 21$ can't be obtained from given data.

The value of

$\log_{2\sqrt{3}}1728$ is equal to

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

Let

$\log _{ { 2\sqrt { 3 } } }{ 1728 } = x$

$\implies 1728 = \left( { 2\sqrt { 3 } } \right) { ^{ x } }\\ \implies (1728 = { 2 }^{ 6 }\times { 3 }^{ 3 } = { 2 }^{ 6 }{ \sqrt { 3 } }^{ 6 })\\ \therefore { 2 }^{ 6 }{ (\sqrt { 3 })}^{ 6 } = { (2\sqrt { 3 } ) }^{ x }\\ \therefore x = 6\\ \\$

$\log_{10}4 = 0.6020$ , then the value of

$\log_{10}8$ is equal to

0%
$0.803$
0%
$0.853$
0%
$0.903$
0%
$0.953$

Explanation

$\log_{10} 8 = \log_{10} (4 \times 2) = \log_{10} 4+ \log_{10} 2$

$=0.6020+0.3010=0.9030$

The value of

$\log_{15}225^{225}$ is equal to

0%
$15$
0%
$225$
0%
$375$
0%
$450$

Explanation

$225\log _{15}{225}$

$=225\log _{15}{{15}^{2}}$

$=225\times 2\log _{15}{15} = 450$

Which of the following is not an equivalent statement for the exponential form

$a^{x} = N$

0%
$x = \sqrt[a]{N}$
0%
$\log_{a}N = x$
0%
$a = \sqrt[x]{N}$
0%
$\log_{x}N = a$

Explanation

Given

${ a }^{ x }=N$

Taking

$\log$ we get

$\log _{ a }{ N } \quad =\quad x$

By logarithmic property,

$N=a^x$

Now taking

$xth$ root of the given equation

$a =\sqrt [ x ]{ N }$

So, Option A and D are not an equivalent statement for the exponential form

$a^x=N$

$3^{\log_{4}{x}}=27$ , then

$x$ is equal to

0%
$16$
0%
$64$
0%
$27$
0%
$\log_2 16$

Explanation

$3^{\log_{4}{x}}=27$

$\Rightarrow { 3 }^{ \log _{ 4 }{ x } }={ 3 }^{ 3 }$
Equating powers of

$3$

$\log _{ 4 }{ x } =3$
Taking exponential of the above equation, we get

${ 4 }^{ 3 }=x$

$\therefore x=64$

The value of

$\log_418$ is equal to

0%
a rational number
0%
an irrational number
0%
a prime number
0%
none of these

Explanation

$log_{4}18$

$=\dfrac{log18}{log4}$

$=\dfrac{log2+log9}{2log2}$

$=\dfrac{log2+2log3}{2log2}$

$=\dfrac{log2}{2log2}+\dfrac{2log3}{2log2}$

$=\dfrac{1}{2}+log_{2}3$

Since

$log_{2}3$ is irrational.

Hence

$\dfrac{1}{2}+log_{2}3$ is irrational.

Therefore

$log_{4}18$ is an irrational number.

Evaluate the expression by using logarithm tables:

$\dfrac{(17.42)^{2/{3}}\times 18.42}{\sqrt{126.37}}$

0%
$11.01$
0%
$12.01$
0%
$13.01$
0%
$14.01$

Explanation

Let

$x= \dfrac{(17.42)^{2/{3}}\times 18.42}{\sqrt{126.37}}$
Taking logarithm on both sides,

$\log { x } =\log { (17.42)^{ 2/{ 3 } } } +\log { 18.42 } -\log { \sqrt { 126.37 } }$

$\log { x } =\dfrac { 2 }{ 3 } \log { 17.42 } +\log { 18.42 } -\dfrac { 1 }{ 2 } \log { (126.37) }$

$\log { x } =\dfrac { 2 }{ 3 } \log { (1.742\times 10) } +\log { (1.842\times 10) } -\dfrac { 1 }{ 2 } \log { (1.264\times { 10 }^{ 2 }) }$

$\log { x } =\dfrac { 2 }{ 3 } { (1.2410) } + { 1.2653 } -\dfrac { 1 }{ 2 } { (2.1018) }$

$\log { x } =0.8273+1.2653-1.0509$

$\log { x } =1.0417$

$\Rightarrow x= \text{antilog }(1.0417)$

$\Rightarrow x = 11.01$

The number

$\displaystyle N= 6 \log_{10}2+\log _{10}31$ lies between two successive integers whose sum is equal to:

0%
5
0%
7
0%
9
0%
10

Explanation

$N=\log _{ 10 }{ 64 } +\log _{ 10 }{ 31 } =\log _{ 10 }{ 1984 }$

Now, we know that

$\log_{10}{1000}=3$ and

$\log_{10}{10000}=4$ .

So,

$1000<1984<10000$

$\Rightarrow\log_{10}{1000}<\log_{10}{1984}<\log_{10}{10000}$

$\Rightarrow3<\log_{10}{1984}<4$

$\Rightarrow3<\log _{ 10 }{ 64 } +\log _{ 10 }{ 31 }<4$

So, it lies between two successive integers

$3$ and

$4$ .

So, their sum is equal to

$3+4=7$

Hence. option

$B$ is correct.

$\log_{5} x = y$ , then value of

$5^{5y}$ is equal to

0%
$\dfrac{x}{5}$
0%
$5x$
0%
$\log_{x}5$
0%
$x^{5}$

Explanation

$\log_{5} x = y \Rightarrow 5^{y} = x$

$\therefore (5^{y})^5 = x^5 \Rightarrow 5^{5y} = x^{5}$

$\log 3 = 0.477$ and

$(1000)^x = 3$ , then the value of

$x$ will be

0%
$0.159$
0%
$0.62$
0%
$0.162$
0%
$0.59$

Explanation

Since,

$(1000)^x=3$

$\therefore \log [(1000)^x] = \log3$

$\Rightarrow x \log 1000 = \log 3$

$\Rightarrow x \log (10^3) = \log 3$

$\Rightarrow x \cdot 3\log 10 = \log 3$

$\Rightarrow 3x = \dfrac {\log 3}{\log 10}$

$\Rightarrow x = \dfrac{0.477}{3 \times 1} = 0.159$

Find the mantissa of the logarithm of the number

$0.002359$ .

0%
$3710$
0%
$3718$
0%
$3728$
0%
$3742$

Explanation

We have to find

$\log (0.002359)$

Firstly, we will write

$0.002359$ in standard form.

So,

$0.002359 = 2.359 \times 10^{-3}$

Here, characteristic is -3.

To find the mantissa of

$\log (0.002359)$ , we first look in the row starting with 23. In this row, look at the number in the column headed by 5. The number is 3711.

Now, move to the column of mean differences and look under the column headed by 9 in the row corresponding to 23. We see the number 17 here.

Add this number to 3711. We get the number 3728. This is the required mantissa of

$\log (0.002359)$ .

Mantissa of

$\log 23.598$ ,

$\log 2.3598$ and 0.023598 is the same (only characteristics are different).

$\displaystyle \log_{10} 8 = 0.90$ , then the value of

$\displaystyle \log_{10}0.125$ is

0%
$0.9$
0%
$1$
0%
$0$
0%
$-0.9$

Explanation

$\log_{10}8 = 0.90$

$\log_{10}0.125 = \log_{10} \dfrac{1}{8} = \log_{10} 8^{(-1)} =-\log_{10}8 = -0.9$ (Using

$\log a^b = b \log a$ )

$\displaystyle \frac {\log 81}{\log 27} = x$ and

$x$ is expressed as

$\displaystyle 1 \frac {1}{m}$ , then

$m$ is equal to

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

Explanation

$\displaystyle \frac {\log 81}{\log 27} = x$

$\displaystyle \Rightarrow x = \dfrac {\log 3^4}{\log 3^3} = \dfrac {4 \log \: 3}{3 \log 3} = \dfrac {4}{3} = 1 \dfrac {1}{3}$

$\therefore m =3$

$\displaystyle \frac {\log 128}{\log 32} = x$ , then the value of

$x$ will be

0%
$\dfrac{5}{7}$
0%
$\dfrac{7}{5}$
0%
$\dfrac{8}{5}$
0%
$\dfrac{5}{8}$

Explanation

Given,

$\displaystyle \frac {\log 128}{\log 32} = x$

$\Rightarrow \dfrac{\log2^7}{\log2^5}=x$

$\Rightarrow \dfrac{7\log2}{5\log2}=x$ (Using

$\log a^b=b\log a$ )

$\therefore x = \dfrac{7}{5}$ .

$\displaystyle \log_x 2 = -1$ , then the value of

$x$ is equal to

0%
$2$
0%
$\dfrac{1}{2}$
0%
$-2$
0%
$1$

Explanation

Since,

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

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

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

$\implies x = \dfrac{1}{2}$ .

$\displaystyle \frac {\log 225}{\log 15} = \log x$ , then the value of

$x$ is equal to

0%
$400$
0%
$300$
0%
$200$
0%
$100$

Explanation

Given,

$\displaystyle \frac {\log 225}{\log 15} = \log x$

$\implies \dfrac{\log 15^2}{\log 15}=\log x$

$\implies \dfrac{2\log 15}{\log 15}=\log x$ (Using

$\log a^b=b\log a$ )

$\implies \log x= 2$

$\implies \log x= \log 100$

$\implies x=100$ .

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

$\displaystyle \log \sqrt {32}$ =

$\cfrac{m}{4}$ , then the value of

$m$ is equal to

0%
$43$
0%
$8$
0%
$3$
0%
$6$

Explanation

Goven,

$\log_{10}8 = 0.9$

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

$\Rightarrow 3\log_{10} = 0.9$ (Using

$\log a^b = b \log a$ )

$\Rightarrow \log_{10}2 = 0.3$
Now,

$\log_{10}\sqrt{32} = \dfrac{5}{2}\log_{10}2 = 2.5\times 0.3 = 0.75= \dfrac{3}{4}$

$\Rightarrow m=3$ .

The value of

$\displaystyle \log_{2} 0.125$ is equal to

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

Explanation

$\log _{ 2 }{ 0.125 } = \log _{ 2 }{ { \dfrac { 1 }{ 8 } } } \\ =\log _{ 2 }{ { 2 }^{ -3 } } \\ = -3$

The value of

$\displaystyle \log (a)^3 \div \log a$ is equal to

0%
$1$
0%
$4$
0%
$6$
0%
$3$

Explanation

Let,

$x = \log (a)^3 \div \log a$

$\therefore x = \dfrac{3\log a}{\log a} = 3$ (Using

$\log a^b = b \log a$ )

$\log_{10} x = 2a$ and

$\displaystyle \log_{10} y = \dfrac {b}{2}$ , then

$\displaystyle 10^{2b + 1}$ in terms of

$y$ is

$\displaystyle my^4$ . Then what is the value of

$m$ ?

0%
$13$
0%
$15$
0%
$10$
0%
$5$

Explanation

Since,

$\log _{ 10 }{ x } = a$ and

$\log _{ 10 }{ y } = \dfrac { b }{ 2 }$

$\implies x = { 10 }^{ a }$ and

$y={ 10 }^{ \dfrac { b }{ 2 } }$

Consider,

${ 10 }^{ 2b+1 }$

$= { 10 }^{ 2b } \cdot 10$

$= { 10 }^{ 4(\dfrac { b }{ 2 } ) } \cdot 10$

$= { y }^{ 4 } \cdot 10$

Hence, C will be correct answer.

The value of

$\displaystyle \log_{3} 27$ is equal to

0%
$3$
0%
$9$
0%
$16$
0%
$25$

Explanation

$\log _{ 3 }{ 27 } = \log _{ 3 }{ { 3 }^{ 3 } } \\ =3\log _{ 3 }{ 3 } \\ = 3$

The value of

$\log_{5} 0.2$ is equal to

0%
$-1$
0%
$1$
0%
$10$
0%
$-10$

Explanation

Let

$\log _{ 5 }{ 0.2 } = x$

$\therefore 0.2 = { 5 }^{ x } \Rightarrow \dfrac { 2 }{ 10 } = { 5 }^{ x }$

$\therefore \dfrac{1}{5} = {5}^{x} \Rightarrow 5^{-1} = 5^{x}$

$\therefore x = -1$

The value of

$\displaystyle \log_{5} 625$ is equal to

0%
$2$
0%
$6$
0%
$4$
0%
$1$

Explanation

$\log _{ 5 }{ 625 } = \log _{ 5 }{ { 5 }^{ 4 } } \\ = 4\log _{ 5 }{ 5 } \\ = 4$

After simplication

$\displaystyle (243)^{-\frac {3}{5}}$ , then answer is

$\displaystyle \frac {1}{27}$
State true or false:

0%
True
0%
False

Explanation

As given

$(243)^{-\frac {3}{5}}$

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

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

$\Rightarrow 3^{-3}$

$\Rightarrow \frac{1}{27}$

$\displaystyle 2 \log y - \log x - 3 = 0$ and

$x$ =

$\dfrac {y^2}{m}$ , then the value of

$m$ is equal to

0%
$4000$
0%
$3000$
0%
$2000$
0%
$1000$

Explanation

Since,

$\displaystyle 2 \log y - \log x - 3 = 0$

$\therefore \log y^2 - \log x = \log 1000$ (

$\log a^b = b \log a$ )

$\Rightarrow \log\dfrac{y^2}{x} = \log 1000$ (

$\log\dfrac{a}{b} = \log a - \log b$ )

$\Rightarrow \dfrac{y^2}{x} = 1000$

$\Rightarrow x = \dfrac{y^2}{1000}$

$\Rightarrow m = 1000$

State true or false.

Solution of

$\displaystyle a^{\tfrac {4}{3}} \div a^{-\tfrac {2}{3}}$ is

$a^{2}$ .

0%
True
0%
False

Explanation

$\displaystyle a^{\tfrac {4}{3}} \div a^{-\tfrac {2}{3}}$

$=\displaystyle \dfrac{a^{\tfrac {4}{3}}} { a^{-\tfrac {2}{3}}}$

$=\displaystyle a^{(\tfrac {4}{3}+\tfrac {2}{3})}$

$=a^{2}$

The given statement is true.

The value of

$\log_{0.5}16$ is equal to

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

Explanation

Given,

$\log _{ 0.5 }{ 16 } = x$

$\implies 16 = {0.5 }^{ x }$

$\implies { 2 }^{ 4 } = { 2 }^{ -x }$

$\implies x = -4$

$\log_{16} 8$ =

$\displaystyle \frac{m}{4}$ , then value of

$m$ is equal to

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

Explanation

Since,

$\log _{16}{8}=\frac {m}{4}$ ...(1)

Let

$\log _{ 16 }{ 8 } =x$

Convert in exponential form,

$8 = { 16 }^{ x }$

${ 2 }^{ 3 } = { 2 }^{ 4x }$

$\therefore x = \displaystyle \frac { 3 }{ 4 }$ ...(2)

From (1) and (2),

$m=3$

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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