MCQ Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 9

Multiply

$10^4$ by

$10^2$

0%
$10^8$
0%
$10^2$
0%
$10^6$
0%
$10^{-2}$
0%
$10^3$

Evaluate using logarithm table:

$\dfrac {28.45 \times \sqrt [3] {0.3254}}{32.43 \times \sqrt [5] {0.3046}}$

0%
$0.7666$
0%
$0.7656$
0%
$0.5686$
0%
$0.2936$

Explanation

Let

$y=\dfrac { 28.45\times \sqrt [ 3 ]{ .3254 } }{ 32.43\times \sqrt [ 3 ]{ .3046 } }$

$\ln { y } =\ln { 28.45 } +\ln { \sqrt [ 3 ]{ .3254 } } -(\ln { 32.43 } +\ln { \sqrt [ 5 ]{ .3046 } } )\\ \ln { y } =\ln { 25.45 } +\dfrac { 1 }{ 3 } \ln { .3245- } \ln { 32.43 } -\dfrac { 1 }{ 5 } \ln { .4046 } \\ \ln { y } =3.236+(-.375)-3.479-(-.237)\\ \ln { y } =-.381$

$y=$ anti

$\ln { (-.381) }$

$y=.7656$

So, option B is correct.

$\dfrac{\log_{2}a}{4} = \dfrac{\log_{2}b}{6} = \dfrac{\log_{2}c}{3p}$ and also

$a^{3}b^{2}c = 1$ , then the value of

$p$ is equal to

0%
$-6$
0%
$-7$
0%
$-8$
0%
$-9$

Explanation

$\dfrac { \log _{ 2 }{ a } }{ 4 } =\dfrac { \log _{ 2 }{ b } }{ 6 } =\dfrac { \log _{ 2 }{ c } }{ 3p } = x\\ \log _{ 2 }{ a } = 4x$

$\Rightarrow a = { 2 }^{ 4x }$

$\log _{ 2 }{ b } = 6x$

$\Rightarrow b = { 2 }^{ 6x }$

$\log _{ 2 }{ c } = 3px$

$\Rightarrow c = { 2 }^{ 3px }$

$\\ { a }^{ 3 }{ b }^{ 2 }c = 1$

${ 2 }^{ 12x }{ \cdot 2 }^{ 12x }\cdot { 2 }^{ 3px } ={ 2 }^{ 0 }$

${ 2 }^{ 24x+3px } ={ 2 }^{ 0 }$

$\therefore 24x +3px = 0$

$\therefore 3px = -24x$

$\therefore p = -8$

Given

$log_3(a) = c$ and

$log_3(b)=2c, a =$

0%
$3c$
0%
$c + 3$
0%
$b^2$
0%
$\sqrt{b}$
0%
$\dfrac{b}{2}$

Explanation

Given,

$\log _{ 3 }{ (a) } =c$ and

$\log _{ 3 }{ (b) } =2c$

$\therefore a={ 3 }^{ c }$ and

$b={ 3 }^{ 2c }$

Now consider

$a^2$ as follows:

$a^{ 2 }={ (3 }^{ c })^{ 2 }$

$\Rightarrow a^{ 2 }={ 3 }^{ 2c }$

$\Rightarrow a^{ 2 }=b$

$\Rightarrow a=\sqrt { b }$

The number

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

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

Explanation

$N=6\log_{10}2+\log_{10}31$

$=\log_{10}2^{6}+log_{10}(31)$

$=\log_{10}(2^{6}.31)$

$=\log_{10}(64\times31)$

$=\log_{10}(1984)$

Now

$\log_{10}(10^{3})=\log_{10}(1000)=3$

Similarly

$\log_{10}(10^{4})=4$

Since

$10^{3}<1984<10^{4}$

$3<\log_{10}(1984)<4$

Hence it lies between

$3$ and

$4$ .

The sum of these two integers is

$7$ .

$\log_{10} 2 = 0.3010$ , then the number of digits in

$2^{64}$ is

0%
$18$
0%
$24$
0%
$22$
0%
$20$

Explanation

Given

$\log _{ 10 }{ 2 } =0.301$

$\log _{ 10 }{ 2^{64} } =64 \times \log _{ 10 }{ 2 } =64 \times 0.3010=19.264$

$\Rightarrow 2^{64}=10^{19.264}$

The number of digits in

$10^{19}$ is

$20$ , there will be

$21$ digits from

$10^{21}$

The number

$10^{19.264}$ lies between them

Therefore the number of digits in

$10^{19.264}$ is

$20$

Therefore the correct option is

$D$

Approximate of

$\log_{11}21$ is

0%
1.27
0%
1.21
0%
1.18
0%
1.15
0%
1.02

Explanation

Approximate value of

$\log _{ 11 }{ 21 }$

$=\log _{ 11 }{ (7\times 3) }$

$=\log _{ 11 }{ 7 } +\log _{ 11 }{ 3 }$

$=0.8115+0.4581$

$=1.27$

$2^{1/4}4^{1/8}8^{1/16}16^{1/32}$ ....... is equal to

0%
1
0%
2
0%
$\frac{3}{2}$
0%
$\frac{5}{2}$

Let

$a = \log_3\log_32$ . An integer k satisfying

$1< 2^{(-k+3^{-a})} < 2,$ must be less than _____.

0%
$1.25766$
0%
$2.256$
0%
$3$
0%
$1$

$log_AD= a,$ then value of

$log_612$ is (in terms of a)

0%
$\frac{1+3a}{3a}$
0%
$\frac{1+2a}{3a}$
0%
$\frac{1+2a}{2a}$
0%
$\frac{1+3a}{2a}$

$x=198!$ then value of the expression

$\dfrac {1}{\log_{2}x}+\dfrac {3}{\log_{2}x}+...\dfrac {198}{\log_{2}x}$ equals ?

0%
$-1$
0%
$0$
0%
$1$
0%
$198$

Given

$log2=a,log3=b$ express the following in terms of

$a$ or

$b$ or both

0%
$\log1.5$
0%
$\log1.2$
0%
$\log0.24$
0%
$\log0.5$
0%
$\log0.036$

$y=a\log\left|x\right|+bx^{2}+x$ has extreme values at

$x=2$ and

$x=-4/3$ then

0%
$a=12,b=-10$
0%
$a=4,b=-3/4$
0%
$a=-6,b=1/4$
0%
$none$

The value of

$\dfrac{log_2 24}{log_{96} 2}-\dfrac{log_2192}{log_{12}{2}}$ is

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

Explanation

Consider

$\dfrac{log_2 24}{log_{96} 2}-\dfrac{log_2192}{log_{12}{2}}\\$

$=\dfrac{log24.log96-log192log12}{(log2)^2}$

$=\dfrac{log(2^3 \times 3)log(2^5\times 3)-log(2^6\times3)log(2^2\times3)}{(log2)^2}$

$=\dfrac{(3log2+log3)(5log2+log3)-(6log2+log3)(2log2+log3)}{(log2)^2}$

$=\dfrac{15(log2)^2-12(log2)^2}{(log2)^2}$

$=3\dfrac{(log2)}{log2}$

$=3$

Option A is the correct answer.

The value of

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

0%
$1$
0%
$2$
0%
$\dfrac{1}{2}$
0%
$4$

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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