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CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 6 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Logarithm And Antilogarithm
Quiz 6
If the value of $$(\log_{10} 2 )+ 1$$ is in the form of $$\log_{10}m$$, then $$m$$ is equal to
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0%
$$11$$
0%
$$14$$
0%
$$20$$
0%
$$17$$
Explanation
$$ \log _{ 10 }{ 2 } +1 = \log _{ 10 }{ 2 } +\log _{ 10 }{ 10 } \\ \\ \\ =\log _{ 10 }{ (2\times 10 } )\\ = \log _{ 10 }{ 20 } $$
The value of $$\displaystyle \log_{10}0.001 $$ is equal to
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0%
$$-3$$
0%
$$3$$
0%
$$-2$$
0%
$$2$$
Explanation
Consider, $$ \log _{ 10 }{ 0.001 } =x$$
$$\implies 0.001 = { 10 }^{ x }$$
$$\implies \dfrac { 1 }{ 1000 } = 10^x$$
$$\implies { 10 }^{ -3 }= 10^x$$
$$\implies x=-3 $$
If $$ \log (a)^{3} + \log a$$ = $$m \log a$$, then value of $$m$$ is equal to
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0%
$$1$$
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$$4$$
0%
$$2$$
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$$-5$$
Explanation
$$\textbf{Step 1 : Use the rule of logarithm}$$
$$\text{According to logarithm power rule,}$$
$$\because \log_e( X ^Y) = Y . \log_e(X)$$
$$\text{Therefore, }\log(a)^3 = 3. \log \:a$$
$$\text{Since,} \log (a)^3 + \log a = m \log a$$
$$\text{Therefore, }3. \log a + \log a=m \log a$$
$$4 \log a=m \:\log\:a$$
$$\text{By comparing both the sides we get,}$$
$$m=4 $$
$$\textbf{Hence, option B is correct.}$$
The logarithm form of $$\displaystyle (81)^\frac{3}{4} = 27$$ is $$\log_{81} 27 = \displaystyle \frac{3}{m}$$. Then value of $$m$$ is equal to
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0%
$$2$$
0%
$$0$$
0%
$$4$$
0%
$$3$$
Explanation
Given,
$$\log_{81} 27 = \displaystyle \frac{3}{m}$$ ...(1)
Now, $$\displaystyle (81)^\frac{3}{4} = 27$$
Converting in logarithm form, we can write
$$\log_{81} 27 = \displaystyle \frac{3}{4}$$. ...(2)
From (1) and (2), we get
$$\displaystyle \dfrac{3}{m}=\displaystyle \dfrac{3}{4}$$
$$\therefore m=4$$
If $$ \dfrac{\log 81}{\log 27} = x$$, then the value of $$x$$ is equal to
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0%
$$\dfrac { 4 }{ 3 } \\ \\ $$
0%
$$\dfrac { 3 }{ 4 } \\ \\ $$
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$$\dfrac { 1 }{ 3 } \\ \\ $$
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Not solvable
Explanation
$$ \dfrac{\log 81}{\log 27} = x$$
$$\Rightarrow x = \dfrac{\log 3^4}{\log 3^3} = \dfrac{4 \log 3}{3 \log 3} = \dfrac{4}{3}$$
The logarithm form of $$10^{-3} = 0.001$$ is $$\log_{10} 0.001 = -m$$, then value of $$m$$ is
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0%
$$-1$$
0%
$$-2$$
0%
$$3$$
0%
$$-4$$
Explanation
Given, $$ { 10 }^{ -3 }= 0.001$$
$$ \implies \log _{ 10 }{ 0.001 } = -3$$
$$\therefore m = 3$$.
If $$\log_{10}(x - 10) = 1$$, then value of $$x$$ is
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0%
$$10$$
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$$13$$
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$$20$$
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$$26$$
Explanation
Since, $$ \log _{ 10 }{ (x-10) } =1$$
$$\implies (x -10)=10^1$$
$$ \implies x = 20 $$.
If $$\displaystyle 64^{a}=\frac{1}{256^{b}}$$, then $$3a + 4b$$ equals:
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$$2$$
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$$4$$
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$$8$$
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$$0$$
Explanation
$$ 64^{a}=\dfrac{1}{256^{b}}$$
$$\Rightarrow (2^{6})^{a}=\dfrac{1}{(2^{8})^{b}}$$
$$ \Rightarrow 2^{6a}\times 2^{8b}=1$$
$$\Rightarrow 2^{6a+8b}=2^{0}$$
$$ \Rightarrow 6a+8b=0$$
$$\Rightarrow 3a+4b=0$$
If $$x = 2$$ and $$y = 3$$, then find the value of $$\left[ \displaystyle\frac { 1 }{ x^{ x } } +\displaystyle\frac { 1 }{ y^{ y } } \right] $$.
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$$ \displaystyle\frac { -31 }{108 } $$
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$$ \displaystyle\frac { 31 }{108 } $$
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$$ \displaystyle\frac { 125 }{171} $$
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$$ \displaystyle\frac { 153 }{222} $$
Explanation
$$\cfrac { 1 }{ x^{ x } } +\cfrac { 1 }{ y^{ y } } =\cfrac { 1 }{ 2^{ 2 } } +\cfrac { 1 }{ 3^{ 3 } }$$
$$ =\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 27 } \\ =\cfrac { 27+4 }{ 108 } $$
$$=\cfrac { 31 }{ 108 } $$
If $$\log_4m\,=\,1.5$$, then the value of $$m$$ is equal to
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$$m=8$$
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$$m=4$$
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$$m=16$$
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$$m=64$$
Explanation
Since, $$\log _{ 4 } m\, =\, 1.5$$
$$\Rightarrow m={ 4 }^{ 1.5 }=2^{2\times 1.5}=2^3=8$$
The value of $$[ 5 (8^{\tfrac 13} + 27^{\tfrac 13} )^3 ]^{\tfrac 14}$$ is
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$$5^4$$
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$$5^{\tfrac 14}$$
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$$5$$
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None of these
Explanation
$$\left [ 5 \left ( 2^{3\times \tfrac {1}{3}}+ 3^{3\times \tfrac {1}{3}} \right )^3 \right ]^{\tfrac 14}$$
$$=[ 5 (2 + 3)^3 ] ^{\tfrac 14}$$
$$= (5 \times 5^3)^{\tfrac 14} = 5^{4\times \tfrac {1}{4}} = 5$$
If $$\displaystyle \sqrt{3^{n}}=81$$. Then n is equal to
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0%
2
0%
4
0%
6
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8
Explanation
$$\Rightarrow \sqrt{3^{n}}=81$$
$$\Rightarrow 3^{n/2}=3^{4}$$
Base are same so
$$\Rightarrow \frac{n}{2}=4\Rightarrow n=8$$
The value of $$512^\frac {-2}{9}$$ is
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$$\displaystyle \frac {1}{2}$$
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$$2$$
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$$4$$
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$$\displaystyle \frac {1}{4}$$
Explanation
$$\begin{aligned}{}{\left( {512} \right)^{ - {\textstyle{2 \over 9}}}} &= {\left( {{2^9}} \right)^{ - {\textstyle{2 \over 9}}}}\\&={2^{9 \times \frac{{ - 2}}{9}}} \quad\quad\quad\because[(a^b)^c=a^{bc}]\\&= {2^{ - 2}}\\ &= \frac{1}{4}\end{aligned}$$
Hence, option $$D$$ is correct.
The value of $$(-3)^0 - (-3)^3 - (-3)^{-1} + (-3)^4 - (-3)^{-2}$$ is
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$$109\, \displaystyle \frac {2}{9}$$
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$$109\, \displaystyle \frac {9}{2}$$
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$$109$$
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None of these
Explanation
Given expression is $$(-3)^0 - (-3)^3 - (-3)^{-1} + (-3)^4 - (-3)^{-2}$$
Using the law of exponent, $$a^{-m}=\dfrac{1}{a^m}$$, we can write it as:
$$=1 - (-27) - \left(-\cfrac {1}{3}\right)$$
$$+ 81 - \cfrac 19$$
$$= 1 + 27 + \cfrac {1}{3} + 81 - \cfrac {1}{9}$$
$$=109 + \cfrac {2}{9}\, $$
$$=\, 109 \displaystyle \frac {2}{9}$$
Hence, option A is correct.
SImplify: $$(256)^{0.16} \times (256)^{0.09}$$
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$$4$$
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$$16$$
0%
$$64$$
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$$256.25$$
Explanation
$$(256)^{0.16} \times (256)^{0.09}$$
$$=(256)^{(0.16 + 0.09)}$$
$$= (256)^{0.25} = (256)^{\tfrac {25}{100}} = (256)^{\tfrac 14}$$
$$= (4^4)^{\tfrac 14} = 4^1 = 4$$
The value of $$4(7^1.\, 7^{-1}.\, 7^{-1}.7^0)$$ is
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0%
$$3\displaystyle \frac {7}{3}$$
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$$\displaystyle \frac {6}{7}$$
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$$3\displaystyle \frac {3}{7}$$
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None of these
The value of $$[10^{150}\, \div\, 10^{146} ]$$ is
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0%
$$1000$$
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$$10000$$
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$$100000$$
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$$10^5$$
Explanation
$$(10)^{150} \div (10)^{146} = \cfrac {(10)^{150}}{(10)^{146}}$$
$$= 10^{(150 - 146)} = 10^4 = 10000$$
$$\log_4 $$1 is equal to
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$$1$$
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$$0$$
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$$\infty$$
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none of these
Explanation
$$log_a 1 = m$$
$$\Rightarrow a^m = 1 a^0 \Rightarrow m = 0$$
Value of $$\displaystyle \log _{4}18 $$ is:
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an irrational number
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a rational number
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natural number
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whole number
Explanation
$$\log_{4}{18}=\log_{2^{2}}{(2.3^{2})}=\dfrac{1}{2}\log_{2}(2.3^{2})$$
$$=\dfrac{1}{2}\left [ \log_{2}{2}+\log_{2}{3^{2}} \right ]=\dfrac{1}{2}\left [ 1+2.\log_{2}{3} \right ]$$
$$\log_{2}{3}$$ is an irrational number.
Hence, $$\dfrac{1}{2}\left [ 1+2.\log_{2}{3} \right ]$$ is also an irrational number.
The value of $$(6^{-1} -7^{-1})^{-1} (5^{-1} -4^{-1})^{-1}$$ is equal to
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0%
$$\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{1}{3}$$
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$$\displaystyle \frac{1}{4}$$
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None of these
Explanation
Given expression is $$\left({ 6 }^{ -1 }-7^{ -1 }\right)^{ -1 }\left(5^{ -1 }-4^{ -1 }\right)^{ -1 }$$.
From laws of exponents, we know that $$a^{-m}=\left(\dfrac{1}{a}\right)^m$$
Hence, the given expression can be simplified as:
$$\left(\cfrac { 1 }{ 6 } -\cfrac { 1 }{ 7 } \right)^{ -1 }\left(\cfrac { 1 }{ 5 } -\cfrac { 1 }{ 4 } \right)^{ -1 }$$
Solving the fractions, we get:
$$\cfrac { 1 }{ 6 } -\cfrac { 1 }{ 7 }=\cfrac{1}{42}$$
and
$$\cfrac { 1 }{ 5 } -\cfrac { 1 }{ 4 }=\cfrac{-1}{20}$$
Hence, the expression simplifies to:
$$\left(\cfrac { 1 }{ 42 } \right)^{ -1 }\left(-\cfrac { 1 }{ 20 } \right)^{ -1 }$$
Again applying the law of exponent, $$a^{-m}=\left(\dfrac{1}{a}\right)^m$$, we get:
$$(42)^1(-20)^1$$
$$\Rightarrow 42\times (-20)=-840$$
Hence, $$\left({ 6 }^{ -1 }-7^{ -1 }\right)^{ -1 }\left(5^{ -1 }-4^{ -1 }\right)^{ -1 }=-840$$.
Value of the expression $$\displaystyle \log _{2}\sqrt[5]{2.\sqrt[3]{2\sqrt{2}}}$$ is
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$$0.1$$
0%
$$0.2$$
0%
$$0.3$$
0%
does not exist
Explanation
$$\log_{2}\left ( \sqrt[5]{2.\sqrt[3]{2\sqrt{2}}} \right )=$$
$$=\log_{2}\left ( \sqrt[5]{2.({2^{3/2}})^{1/3}} \right )$$
$$=\log_{2}\left ( \sqrt[5]{2.2^{1/2}} \right )$$
$$=\log_{2}\left ( 2^{3/2} \right )^{1/5}=\dfrac{1}{5}\, \log_{2}.{2^{3/2}}$$
$$=\dfrac{3}{10}.\log_{2}{2}=\dfrac{3}{10}=0.3$$
Simplify : $$(6^{-1} - 8^{-1})^{-1} + (2^{-1} - 3^{-1})^{-1}$$ is-
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$$20$$
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$$30$$
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$$10$$
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$$40$$
Explanation
Law of exponents,
Given - $$(6^{-1}-8^{-1})^{-1}+(2^{-1}-3^{-1})^{-1}$$
$$=\left(\dfrac{1}{6}-\dfrac{1}{8}\right)^{-1}+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)^{-1}$$
$$=\left(\dfrac{8-6}{6\times 8}\right)^{-1}+\left(\dfrac{3-2}{2\times 3}\right)^{-1}$$
$$=\left(\dfrac{2}{36\times 8}\right)^{-1}+\left(\dfrac{1}{2\times 3}\right)^{-1}$$
$$=\left(\dfrac{1}{24}\right)^{-1}+\left(\dfrac{1}{6}\right)^{-1}$$
$$=24+6$$
$$=30$$
Hence, option B is correct.
If $$\left (\dfrac {a}{b}\right )^{x-1}=\left (\dfrac {a}{b}\right )^{x-3}$$ then the value of $$x$$ is
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-1
0%
1
0%
2
0%
3
Explanation
Given,
$$\left (\dfrac {a}{b}\right )^{x-1}=\left (\dfrac {a}{b}\right )^{-(x-3)}$$
as the terms are same we can equate the powers/exponents of the terms
$$\Rightarrow x-1=-(x-3)$$
$$\Rightarrow x-1=-x+3$$
$$\Rightarrow 2x=4$$
$$\Rightarrow x=2$$
If $$x=\log _{ a }{ bc } ,y=\log _{ b }{ ca } ,z=\log _{ c }{ ab } $$, then the value of $$\dfrac { 1 }{ 1+x } +\dfrac { 1 }{ 1+y } +\dfrac { 1 }{ 1+z } $$ will be
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0%
$$x+y+z$$
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$$1$$
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$$ab+bc+ca$$
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$$abc$$
Explanation
Now, $$1+x=\log _{ a }{ a } +\log _{ a }{ bc } =\log _{ a }{ abc } $$
$$\Rightarrow \dfrac { 1 }{ 1+x } =\log _{ abc }{ a } $$
Similarly, $$\dfrac { 1 }{ 1+y } =\log _{ abc }{ b } $$ and $$\dfrac { 1 }{ 1+z } =\log _{ abc }{ c } $$
$$\therefore \dfrac { 1 }{ 1+x } +\dfrac { 1 }{ 1+y } +\dfrac { 1 }{ 1+z } $$
$$=\log _{ abc }{ a } +\log _{ abc }{ b } +\log _{ abc }{ c } $$
$$=\log _{ abc }{ abc } =1$$
If $$\displaystyle { log }_{ 5 }{ log }_{ 5 }{ log }_{ 2 }x=0$$, then the value of $$x$$ is
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0%
$$32$$
0%
$$125$$
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$$625$$
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$$25$$
Explanation
Given, $$\displaystyle { log }_{ 5 }{ log }_{ 5 }{ log }_{ 2 }x=0$$
$$\displaystyle \Rightarrow \quad { log }_{ 5 }{ log }_{ 2 }x=50$$
$$\displaystyle \Rightarrow \quad { log }_{ 5 }{ log }_{ 2 }x=1\Rightarrow { log }_{ 2 }x=5$$
$$\displaystyle \Rightarrow x={ 2 }^{ 5 }=32$$
The value of $$\displaystyle \log_{2}\left [ \log_{2}\left \{ \log_{3}\left ( \log_{3}27^{3} \right ) \right \} \right ]$$ is equal to
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0%
2
0%
3
0%
0
0%
1
Explanation
Required value of $$\displaystyle \log _{2}\left [ \log _{2}\left \{ \log _{3}\left ( \log _{3}27^{3} \right ) \right \} \right ]$$
$$=$$ $$\displaystyle \log _{2}\left [ \log _{2}\left \{ \log _{3}\left ( \log _{3}3^{9} \right ) \right \} \right ]$$
$$=$$ $$\displaystyle \log _{2}\left [ \log _{2}\left \{ \log _{3}\left (9\cdot \log _{3}3 \right ) \right \} \right ]$$
$$=$$ $$\displaystyle \log _{2}\left [ \log _{2} \left \{\log _{3}\cdot 9 \right \}\right ]$$ ....$$\displaystyle \left [ \because \log _{3}3=1 \right ]$$
$$=$$ $$\displaystyle \log _{2} \left [ \log _{2}\left \{ \log _{3}3^{2} \right \} \right ]$$
$$=$$ $$\displaystyle \log _{2} \left [ \log _{2} 2\cdot \log _{3} 3\right ]$$
$$=$$ $$\displaystyle \log _{2}\left [ \log _{2}2\right ]$$ ....$$\displaystyle \left [ \because \log _{2}2=1 \right ]$$
$$=$$ $$\displaystyle \log _{2}$$ $$1 = 0$$
If there are $$n$$ zeros after the decimal point, then the characteristic of that number will be
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0%
$$n+1$$
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$$-n+1$$
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$$-(n+1)$$
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$$n-1$$
Explanation
For number less than $$1,$$ if there are $$n$$ zeroes after the decimal point, then the characteristics of that number will be $$-(n+1).$$
and For number greater than $$1,$$ if there are $$n$$ number of zeroes are on the left sides of the digits then characteristics will be $$(n+1).$$
Hence, C is the correct option.
The characteristic of a number having $$m$$ $$(m>1)$$ digits is given by,
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$$m-1$$
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$$m+1$$
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$$m$$
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None of the above
Explanation
If a number is $$N>0$$, t
hen $$\log_{10}N$$ will have two parts, the integral part is known the characteristic and the decimal part is known as mantissa.
$$2$$ digit belongs from $$[10,100]$$ where $$\log_{10}10=1$$ and $$\log_{10}100=2$$
Similarly, $$m$$ digit number belongs from $$[10^{m-1},10^m]$$, where $$\log_{10}10^{m-1}=m-1$$ and $$\log_{10}10^m=m$$.
Thus any number between
$$[10^{m-1},10^m]$$ will have $$m-1$$ as the integral part.
Thus the characteristic of a number having $$m$$ digits is given by $$m-1$$.
Let $$x = (0.15)^{20}$$. Find the characteristic in the logarithm of $$x$$ to the base $$10$$.
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$$17$$
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$$21$$
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$$-21$$
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$$-17$$
Explanation
Given $$x=(0.15)^{20}$$
By applying $$\log$$ on both sides , we get $$\log _{ 10 }{ x } =20\log _{ 10 }{ (0.15) } =-16.478$$
$$\Rightarrow \log _{ 10 }{ x } =-17+0.5218$$
The integral part of $$\log _{ 10 }{ x } $$ is called characteristic
Therefore the characteristic of given number is $$-17$$
So option $$D$$ is correct
The value of $$\log_{10} 0.0006024$$ is equal to
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$$\overline {3}.7979$$
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$$\overline {1}.9779$$
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$$\overline {4}.7799$$
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$$0.7279$$
Explanation
$$\log_{10}0.0006024$$
Characteristics$$=-4$$
For mantissa ,read from the table $$6024.$$
Mantissa$$=7799$$
Thus, $$\log_{10}0.0006024=$$ characteristics of $$0.0006024+$$ mantissa of $$0.0006024$$
$$=-4+0.7799$$
$$=\bar4.7799$$
Hence, C is the correct option.
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