MCQ Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 8

$a=\log_35$ and

$b= \log_725$ then correct option is:

0%
$a < b$
0%
$a > b$
0%
$a= b$
0%
None of these

Explanation

$a=\log _{ 3 }{ 5 } =\cfrac { \log { 5 } }{ \log { 3 } } ,b=\cfrac { \log { 25 } }{ \log { 7 } } =\cfrac { \log { 5^2 } }{ \log { 7 } }=\cfrac { 2\log { 5 } }{ \log { 7 } }$

$\cfrac { a }{ b } =\cfrac { \log { 5 } }{ \log { 3 } } \times \cfrac { \log { 7 } }{ 2\log { 5 } } =\cfrac { 1 }{ 2 } \log _{ 3 }{ 7 } =\log _{ 3 }{ (\sqrt { 7 } ) }$

$Now,\sqrt { 7 } <3,so\quad \cfrac { a }{ b } <1$

$\cfrac { a }{ b } <1\ =>a<b$

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

0%
$\dfrac{5}{2}$
0%
$\dfrac{2}{5}$
0%
$25$
0%
$5$

Explanation

$\\(\frac{8^{-1}\cdot 5^3}{2^{-4}\cdot 625})\\= (\frac{2^4\cdot 5^3}{8\cdot 5^4})\\= (\frac{16}{8\cdot 5})\\=(\frac{2}{5})$

Multiple Correct:

Which of the following statements are true

0%
$\log _{ 2 }{ 3 } <\log _{ 12 }{ 10 }$
0%
$\log _{ 6 }{ 5 } <\log _{ 7 }{ 8 }$
0%
$\log _{ 3 }{ 26 } <\log _{ 2 }{ 9 }$
0%
$\log _{ 16 }{ 15 } >\log _{ 10 }{ 11 } >\log _{ 7 }{ 6 }$

If the approximate value of

$\log _{ 10 }{ (4.04) }$ is

$\ abcdef,$ it is given that

$\log _{ 10 }{ (4) }=0.6021$ &

$\log _{ 10 }{ (e) }=0.4343,$ then the value of abcd must be

0%
$6064$
0%
$6063$
0%
$6065$
0%
N.O.T

$\log_{10}e=0.4343$ , then

$\log_{10}1016$ is

0%
$2.99$
0%
$3$
0%
$3.006949$
0%
$3.02$

Explanation

$\log_{10}1016\Rightarrow \dfrac{\log 1016}{\log 10}$

$\Rightarrow \dfrac{3.006949}{1}$

$\rightarrow$ Option

$C$ is correct

The number of zeroes after decimal and before first significant digit in

$(50)^{-100}$ is equal to : (take

$log_{10}$ 5=0.699)

0%
168
0%
169
0%
170
0%
171

Find the value of

$\log_{10}{\left(0.\bar{9}\right)}$

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

Explanation

To find value of

$\log_{10}{0.\bar9}$

Let

$x=0.\bar{9}=0.999999...$

$\Rightarrow 10x=9.99999....$

$\Rightarrow 10x-x=9$

$\Rightarrow 9x=9$

$\Rightarrow x=\dfrac{9}{9}=1$

$\therefore x=1$

Let

$y=\log_{10}{1}$

$\Rightarrow 1={10}^{y}$

$\Rightarrow {10}^{y}={10}^{0}$ (since

${10}^{0}=1$ )

Since bases are same we can equate the powers

$\therefore y=0$

Hence,

$\log_{10}{\left(0.\bar{9}\right)}=0$

$\log 4=1.3868$ , then the approximate value of

$\log\, (4.01)$

0%
$1.3968$
0%
$1.3898$
0%
$1.3893$
0%
$1.9338$

Explanation

Let

$y=f(x)=\log x$

Let

$x=4$

$X+\triangle x=4.01$

$\triangle x=0.01$

For

$x=4$

$Y=\log4=1.3868$

$y=\log x$

$\dfrac{dy}{dx}=\dfrac{1}{x}=\dfrac{1}{4}$

$\triangle y=dy$

$=\dfrac{dy}{dx}.dx$

$=\dfrac{1}{4}\times 0.01$

$\triangle y=0.0025$

$\log(4.01)=y+\triangle y$

$=1.3893$

The equation

$\log_{e}x+\log_{e}(1+x)=0$ can be written as

0%
$x^{2}+x-e=0$
0%
$x^{2}+x-1=0$
0%
$x^{2}+x+1=0$
0%
$x^{2}+xe-e=0$

Explanation

$\log _{ e }{ x } +\log _{ e }{ \left( 1+x \right) } =0$

$\log _{ e }{ x\left( 1+x \right) } =0$

$x\left( x+1 \right) =0$

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

${ x }^{ 2 }+x-1=0$

The logarithmic form of

$4 = 2^2$ is

0%
$log_{ 2 }^{ 4 }= 2$
0%
$log_{ 2 }^{ 2 }= 2$
0%
$log_{ 2 }^{ 4 }=4$
0%
None of these

Explanation

Since,

$4=2^2$

$\implies \log$ of

$4$ to base

$2$ is

$2$

$\implies \log_{2}2= {4}$ .

$x=500,y=100$ and

$z=5050$ , then the value of

$(\log _{ xyz }{ { x }^{ z } } )(1+\log _{ x }{ yz } )$ is equal to.

0%
500
0%
100
0%
5050
0%
10

Explanation

Given,

$\left(\log _{xyz}\left(x^z\right)\right)\left(1+\log _x\left(yz\right)\right)$

$=z\log _{xzy}\left(x\right)\left(\log _x\left(zy\right)+1\right)$

from given w have,

$=5050\log _{(500 \times 5050 \times 100)}\left(500\right)\left(\log _{500}\left((5050 \times 100)\right)+1\right)$

$=\dfrac{5050\log_e \left(505000\right)}{\log_e \left(252500000\right)}+5050\log _{252500000}\left(500\right)$

$=5050$

The greatest value of

$(4\log_{10}{x}-\log_{x}{(0.0001)})$ for

$0 < x < 1$ is

0%
$4$
0%
$-4$
0%
$8$
0%
$-8$

Explanation

$(4log_{10}x-log_x(0.0001)$

$(4log_{10}x-log_{x}\dfrac{1}{10^4}$

$(4log_{10}x-(log_x1-log_x{10^4})$

$(4log_{10}x-0+log_x{10^4})$

$(4log_{10}x+log_x{10^4})$

$4(log_{10}x+log_x10)$

$4(log_{10}x+\dfrac{1}{log_{10}x})$

$AM\geq GM$

$\dfrac{x+y}{2} \geq \sqrt{xy}$

$log_{10}x+\dfrac{1}{log_{10}x} \geq 2$

$4(log_{10}x+\dfrac{1}{log_{10}x}) \geq 8$

The value of

$3 ^{log_4 5} -5 ^{log_4 3}$

0%
$0$
0%
$1$
0%
$2$
0%
None of these

The number of

$\log_2 7$ is

0%
an integer
0%
a rational number
0%
an irrational number
0%
a prime number

The value of

$0.2^{log_{\sqrt{5}} \Big( \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dots \Big)}$ is

0%
$4$
0%
$log 4$
0%
$log 2$
0%
none of these

Explanation

$\textbf{Given that:}$

$0.2^{\log_{\sqrt{5}} \left (\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + ... \right )}$

$....(1)$

$\textbf{Step 1: Let}\ y = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + ...$

This is a infinite term G.P

where

$a = \dfrac{1}{4} , r = \dfrac{1}{2} , \left [S_{y} = \dfrac{a}{1 - r} \right ]$

Hence sum of

$S_{y} = \dfrac{\dfrac{1}{4}}{1 - \dfrac{1}{2}}$

$S_{y} = \dfrac{\dfrac{1}{4}}{\dfrac{1}{2}} = \dfrac{1}{2}$

$....(2)$

Put the this value in eq.

$(1)$

$= (0.2)^{\log_{\sqrt{5}} \left (\dfrac{1}{2} \right )}$

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

$....(3)$

$\textbf{Step 2: Using the principle properties of logarithm}$

$\Rightarrow \log_{a^{P}} {m} = \dfrac{1}{p} \log_{a} m$ [

$a>0,\neq1 , m>0$ ]

$\Rightarrow a^{\log_{a} m} = m$

So from eq.

$(3)$

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

$=(5^{-1})^{2 \log_{5} \frac{1}{2}}$

$=5^{-2 \log_{5} \frac{1}{2}}$

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

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

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

$= 2^{2}$

$= 4$

The value of

$\frac { \log _{ 2 }{ 24 } }{ \log _{ 96 }{ 2 } } -\frac { \log _{ 2 }{ 192 } }{ \log _{ 12 }{ 2 } }$ is:

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

Explanation

Let

$\log _{ 2 }{ 12 }=a$ , then

$\frac { 1 }{ \log _{ 96 }{ 2 } }=\log _{ 2 }{ 96 }=\log _{ 2 }{ { 2 }^{ 3 } } \times 12=3+a$

$\log _{ 2 }{ 24 }=1+a$

$\Rightarrow \log _{ 2 }{ 192 }=\log _{ 2 }{ \left( 16 \times 12 \right) }=4+a$ and

$\frac { 1 }{ \log _{ 12 }{ 2 } } =\log _{ 2 }{ 12 } =a$

Therefore, the given expression =

$(1+a)(3+a)-(4+a)a=3$ .

$(-4)^{4}\times (4)^{1}=(4)^{5}$

0%
True
0%
False

Explanation

False

$LHS=(-4)^{4}\times (4)^{1}$
Using law of exponents

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

$(-4)^{4}\times (4)^{1}=(4)^{4+1} =(4)^{5}$

$\left(\dfrac{2}{3}\right)^{2}\times \left(\dfrac{2}{3}\right)^{5}=\left(\dfrac{2}{3}\right)^{10}$

0%
True
0%
False

Explanation

False.

$LHS=\left(\dfrac{2}{3}\right)^{2}\times \left(\dfrac{2}{3}\right)^{5}$
Using law of exponents,

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

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

$LHS\neq RHS$

$a\times a\times b\times b\times b$ can be written as

0%
$a^2b^3$
0%
$a^3b^2$
0%
$a^3b^3$
0%
$a^5b^5$

Explanation

we have,

$a\times a\times b\times b\times b$

$=a^2 \times b^3 =a^2b^3$

Hence,

$a\times a\times b\times b\times b$ can be written as

$a^2b^3$

$\left(-\dfrac{8}{2}\right)^{0}=0$

0%
True
0%
False

Explanation

$LHS=\left(-\dfrac{8}{2}\right)^{0}$
Using law of exponent

$=a^{0}=1$

$\Rightarrow \left(-\dfrac{8}{2}\right)^{0}=1$

$LHS\neq RHS$

$(-7)^{4}\times (-7)^{2}=(-7)^{6}$

0%
True
0%
False

Explanation

True.

$LHS=(-7)^{4}\times (-7)^{2}$
Using law exponents,

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

$(-7)^{4}\times (-7)^{2}=(-7)^{4+2}=(-7)^{6}$

$LHS=RHS$

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

0%
True
0%
False

Explanation

False

$LHS=(-5)^{2}\times (-5)^{3}$
Using law of exponents,

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

$(-5)^{2}\times (-5)^{3}=(-5)^{2+3}=(-5)^{5}$

$\left(\dfrac {1}{10}\right)^0$ is equal to

0%
$0$
0%
$\dfrac {1}{10}$
0%
$1$
0%
$10$

Explanation

Using law of exponents,

$a^0=1$
where

$a\neq 1$

$\Rightarrow \ \left(\dfrac {1}{10}\right)^0 =1$

By solving

$(6^0 -7^0) \times (6^0+7^0)$ , we get ________.

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

Explanation

$0$

$(6^0 -7^0) \times (6^0+7^0)$

$=(1-1) \times (1+1)$

$=0\times 2=0$

$x$ be any integer different from zero and

$m,n$ be any integers then

$({x^m})^n$ is equal

0%
$x^{m+n}$
0%
$x^{mn}$
0%
$\dfrac {m}{x^n}$
0%
$x^{m-n}$

Explanation

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

[Where,

$a$ is non-zero integer]

Similarly,

$(x^m)^n =(x)^{m\times n}=(x)^{mn}$

State whether the following statement is true (T) or false (F):

$5^0 \times 3^0 = 8^0$

0%
True
0%
False

Explanation

$5^0 \times 3^0 = 8^0$

$1 \times 1 =1$ (

$\because 5^0=1, 3^0 = 1, 8^0 =1$ )

$\log_{\sqrt{2}} x = 4$ then value of

$x$ will be

0%
$4\sqrt{2}$
0%
$\frac{1}{4}$
0%
$4$
0%
$4*\sqrt{2}$

Explanation

$\because \log_{a}n=x$

$\therefore a^{x}=n$
Given

$\log_{\sqrt{2}} x= 4$
Then

$(\sqrt{2})^{4} = x$

$\Rightarrow x = (\sqrt{2})^{4} = (2^{1/2})^{4} = 2^{2} = 4$
Hence, option (C) is correct.

The value of

$\log (1 + 2 * 3)$ :

0%
$2\log3$
0%
$\log1.\log.2+\log3$
0%
$\log1+\log2+\log3$
0%
$\log7$

Explanation

$\log (1 + 2 * 3) = \log (1 + 6) = \log 7$
Hence, option (D) is correct

Number

$\log_{2} 7$ is:

0%
Integer
0%
Rational
0%
Irrational
0%
Prime

Explanation

$\log_{2} 7 = x$

$7 = 2^{x}$
Taking

$\log$ on both sides,

$\log 7 = \log 2^{x}$

$\Rightarrow x \log 2 = \log 7$

$\Rightarrow x=\dfrac{\log 7}{\log 2}=\dfrac{0.8451}{0.3010}=2.8076$
= Irrational number.
Hence, option (C) is correct.

$\log_{x} 243 = 2.5$ , then value of

$x$ will be:

0%
$9$
0%
$3$
0%
$1$
0%
$81$

Explanation

$\because \log_{a} n = x$

$\Rightarrow a^{x} = n$
Similarly

$\log_{x} 243 = 2.5$

$243 = x^{2.5}$

$x^{5/2} = 243 = (3)^{5}$
Squaring on both sides,

$(x^{5/2})^{2} = [(3)^{5}]^{2}$

$x^{5} = (3^{2})^{5}$
On comparing,

$x = 3^{2} = 9$
Hence, option (A) is correct

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 8 - 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 8 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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