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CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 9 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Logarithm And Antilogarithm
Quiz 9
Multiply $$10^4$$ by $$10^2$$
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$$10^8$$
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$$10^2$$
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$$10^6$$
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$$10^{-2}$$
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$$10^3$$
Evaluate using logarithm table:
$$\dfrac {28.45 \times \sqrt [3] {0.3254}}{32.43 \times \sqrt [5] {0.3046}}$$
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$$0.7666$$
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$$0.7656$$
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$$0.5686$$
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$$0.2936$$
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.
If $$\dfrac{\log_{2}a}{4} = \dfrac{\log_{2}b}{6} = \dfrac{\log_{2}c}{3p}$$ and also $$a^{3}b^{2}c = 1$$, then the value of $$p$$ is equal to
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$$-6$$
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$$-7$$
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$$-8$$
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$$-9$$
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 $$log_3(b)=2c, a =$$
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$$3c$$
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$$c + 3$$
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$$b^2$$
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$$\sqrt{b}$$
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$$\dfrac{b}{2}$$
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
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$$5$$
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$$7$$
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$$9$$
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$$10$$
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 $$7$$.
If $$\log_{10} 2 = 0.3010$$, then the number of digits in $$2^{64}$$ is
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$$18$$
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$$24$$
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$$22$$
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$$20$$
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
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1.27
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1.21
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1.18
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1.15
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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
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1
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2
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$$\frac{3}{2}$$
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$$\frac{5}{2}$$
Let $$a = \log_3\log_32$$. An integer k satisfying $$1< 2^{(-k+3^{-a})} < 2,$$ must be less than _____.
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$$1.25766$$
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$$2.256$$
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$$3$$
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$$1$$
If $$log_AD= a, $$ then value of $$log_612$$ is (in terms of a)
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$$\frac{1+3a}{3a}$$
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$$\frac{1+2a}{3a}$$
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$$\frac{1+2a}{2a}$$
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$$\frac{1+3a}{2a}$$
If $$x=198!$$ then value of the expression $$\dfrac {1}{\log_{2}x}+\dfrac {3}{\log_{2}x}+...\dfrac {198}{\log_{2}x}$$ equals ?
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$$-1$$
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$$0$$
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$$1$$
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$$198$$
Given $$log2=a,log3=b$$ express the following in terms of $$a$$ or $$b$$ or both
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$$\log1.5$$
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$$ \log1.2$$
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$$\log0.24$$
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$$ \log0.5$$
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$$\log0.036$$
If $$y=a\log\left|x\right|+bx^{2}+x$$ has extreme values at $$x=2$$ and $$x=-4/3$$ then
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$$a=12,b=-10$$
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$$a=4,b=-3/4$$
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$$a=-6,b=1/4$$
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$$none$$
The value of $$\dfrac{log_2 24}{log_{96} 2}-\dfrac{log_2192}{log_{12}{2}}$$ is
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$$3$$
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$$0$$
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$$2$$
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$$1$$
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
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$$1$$
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$$2$$
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$$\dfrac{1}{2}$$
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$$4$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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