MCQ Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 2

The value of

$(\sqrt 8)^{\tfrac 13}$ is

0%
$2$
0%
$4$
0%
$\sqrt 2$
0%
$8$

Explanation

$(\sqrt 8)^{\tfrac 13} = (8^{\tfrac 12})^{\tfrac 13}$

$= 8^{ ( \tfrac {1}{2} \times \tfrac {1}{3} )} = 8^{\tfrac {1}{6}}$

$\Rightarrow (2^3)^{\tfrac 16} = 2^{(3\times \tfrac {1}{6})} = 2^{\tfrac {1}{2}} = \sqrt 2$

Simplify

$(32)^{\displaystyle \frac {-2}{5}}\, \div\, (125)^{\displaystyle \frac {-2}{3}}$

0%
$\displaystyle \frac {4}{25}$
0%
$\displaystyle \frac {25}{4}$
0%
$\displaystyle \frac {2}{5}$
0%
$\displaystyle \frac {5}{2}$

Explanation

$\cfrac {(32)^{ \tfrac {-2}{5}}}{(125)^{ \tfrac {-2}{3}}} = \cfrac { \cfrac {1}{4}}{ \cfrac {1}{25}} = \cfrac {25}{4}$

If x = 2, y = 3 then

$\displaystyle \frac {1}{x^y}\, +\, \displaystyle \frac {1}{y^x}$ = .........

0%
72
0%
$\displaystyle \frac {17}{72}$
0%
$\displaystyle \frac {31}{108}$
0%
None of these

Explanation

$\displaystyle \frac {1}{2^3}\, +\, \displaystyle \frac {1}{3^2}\, =\, \displaystyle \frac {1}{8}\, +\, \displaystyle \frac {1}{9}\, =\, \displaystyle \frac {9+8}{72}\, =\, \displaystyle \frac {17}{72}$

$\left ( \displaystyle \frac {16}{81} \right )^{\displaystyle \frac {3}{4}}$ = ..........

0%
$\displaystyle \frac {9}{2}$
0%
$\displaystyle \frac {2}{9}$
0%
$\displaystyle \frac {8}{27}$
0%
$\displaystyle \frac {27}{8}$

Explanation

$\left ( \cfrac {16}{81} \right )^{ \tfrac {3}{4}} = \left [ \left ( \cfrac {2}{3} \right )^4 \right ]^{\tfrac {3}{4}} = \cfrac {8}{27}$

The value of

$(3^{0} - 4^{0})\, \times\, 5^2$ is

0%
$25$
0%
$0$
0%

$-25$
0%
none of these

Explanation

To find the value of

$(3^{0} - 4^{0})\, \times\, 5^2$

$(3^{0} - 4^{0}) \times 5^2$

$= (1 - 1) \times 25$

$= 0\times25$

$= 0$

Hence, the value is

$0$

$(64)^{\displaystyle \frac {-2}{3}}\, \times \left ( \displaystyle \frac {1}{4} \right )^{-3}$ equals to

0%
4
0%
$\displaystyle \frac {1}{4}$
0%
1
0%
16

Explanation

$(64)^{ -\tfrac {2}{3}} \times \left ( \cfrac {1}{4} \right )^{-3} = (4^3)^{ -\tfrac {2}{3}} \times \left ( \cfrac {1}{4} \right )^{-3}$

$\Rightarrow 4^{-2} \times \cfrac {1}{4^{-3}}$

$\Rightarrow 4^{-2+3} = 4^1 = 4$

The value of

$(256)^{\dfrac 54}$ is

0%
$512$
0%
$984$
0%
$1024$
0%
$1032$

Explanation

$(256)^{\dfrac 54} = (4^4)^{\dfrac 54}$

$= 4^{(4\times \dfrac {5}{4})}$

$= 4^5 = 1024$

What is the value of

$2^{0.64}*2^{0.36}$ ?

0%
2
0%
1
0%
16
0%
32

Explanation

$2^{0.64}*2^{0.36} =2^{0.64+0.36} = 2^{1.00} = 2^1=2\\ (\because a^m*a^n=a^{m+n})$

The value of

$[ (-2)^{(-2)} ]^{(-3)}$ is

0%
64
0%
32
0%
cannot be determined
0%
none of these

Explanation

$[ (-2)^{(-2)} ]^{(-3)}\, =\, (-2)^6\, =\, 64$

Which of the following values are equal?
I.

$1^4$

II.

$4^0$

III.

$0^4$

IV.

$4^1$

0%
I and II
0%
II and III
0%
I and III
0%
I and IV

Explanation

$1^4 = 1$

II:

$4^0 = 1$

III:

$0^4 = 0$

IV:

$4^1 = 4$

Hence,

$1^4=4^0 = 1$

Evaluate:

$(256)^{0.16}\, \times\, (256)^{0.09}$

0%
$4$
0%
$16$
0%
$64$
0%
$256.25$

Explanation

${ \left( 256 \right) }^{ 0.16 }\times { \left( 256 \right) }^{ 0.09 }$

$={ \left( 256 \right) }^{ 0.16+0.09 }$

$={ \left( 256 \right) }^{ 0.25 }$

$={ \left( 256 \right) }^{ \tfrac { 25 }{ 100 } }$

$={ \left( 256 \right) }^{ \tfrac { 1 }{ 4 } }$

$={ 2 }^{ 8\times \tfrac { 1 }{ 4 } }$

$={ 4 }$

What is the value of

$[\log_{10} (5\log_{10} 100)]^{2}$ ?

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

Explanation

The value of

$[\log_{10}(5\log_{10} 100)]^{2}$ is

$=[\log_{10}(5\log_{10}10^2)]^2$

$=[\log_{10}(10\log_{10}10)]^2$ .....As

$\log a^m=m\log a$

$=[\log_{10}(10\times 1)]^2$ ....As

$\log_aa=1$

$=[\log_{10}10]^2$

$=[1]^2=1$

The value of

$\displaystyle \frac {1}{(216)^{-2/3}}\, +\, \displaystyle \frac {1}{(256)^{-3/4}}\, +\, \displaystyle \frac {1}{(32)^{-1/5}}$ is:

0%
$102$
0%
$105$
0%
$107$
0%
$109$

Explanation

$\cfrac {1}{(216)^{-\tfrac 23}} + \cfrac {1}{(256)^{-\tfrac 34}} + \cfrac {1}{(32)^{-\tfrac 15}}$

$= \cfrac {1}{(6^3)^{-\tfrac 23}} + \cfrac {1}{(4^4)^{-\tfrac 34}} + \cfrac {1}{(2^5)^{-\tfrac 15}}$

$= \cfrac {1}{6^{-2}} + \cfrac {1}{4^{-3}} + \cfrac {1}{2^{-1}}$

$= (6^2 + 4^3 + 2^1)$

$= (36 + 64 + 2^1)$

$= 102$

The value of

$(8^{-25}\, -\, 8^{-26})$ is

0%
$7\, \times\, 8^{-25}$
0%
$7\, \times\, 8^{-26}$
0%
$8\, \times\, 8^{-26}$
0%
None of these

Explanation

$(8^{-25} - 8^{-26})$

$=\left ( { \dfrac {1}{8^{25}}} - { \dfrac {1}{8^{26}}} \right )$

$=\dfrac{8^{25}(8-1)}{8^{25}\times 8^{26}}$

$= \cfrac {(8-1)}{8^{26}}$

$=7 \times 8^{-26}$

Thus option

$(B)$ is correct.

The value of

$\left ( \displaystyle \frac {-1}{216} \right )^{-2/3}$ is

0%
$36$
0%
$-36$
0%
$\dfrac {1}{36}$
0%
$-\dfrac{1}{36}$

Explanation

$\left (\cfrac {-1}{216} \right )^{-\tfrac 23} = \left [ \left (\cfrac {-1}{6} \right )^3 \right ]^{-\tfrac23}$

$=\left ( \cfrac {-1}{6} \right )^{3 \times -\tfrac {2}{3}}$

$=\left (- \cfrac {1}{6} \right )^{-2}$

$= \cfrac {1}{(-1/6)^2}$

$= \cfrac {1}{(1/36)} = 36$

Evaluate:

$(0.04)^{-1.5}$

0%
$25$
0%
$125$
0%
$250$
0%
$625$

Explanation

$(0.04)^{-1.5} = \left ( \cfrac {4}{100} \right )^{-1.5}$

$=\,\left (\cfrac {1}{25} \right )^{-\tfrac 32}$

$= (25)^{\tfrac 32}$

$= (5^2)^{\tfrac 32}$

$= 5^3 = 125$

$\displaystyle 2^{73}-2^{72}-2^{71}$ is the same as :

0%
$\displaystyle 2^{69}$
0%
$\displaystyle 2^{70}$
0%
$\displaystyle 2^{71}$
0%
$\displaystyle 2^{72}$

Explanation

$2^{73}-2^{72}-2^{71}=2^{71}(2^2-2^1-1)$

$=2^{71}(4-2-1)$

$=2^{71}$

Therefore, Option C is correct.

$\dfrac{(23)^{9}-(23)^{8}}{22}=(23)^{x}$ then

$x$ equals :

0%
$7$
0%
$8$
0%
$9$
0%
$1$

Explanation

$LHS= \dfrac{(23)^{9}-(23)^{8}}{22}$

$=\dfrac{23^8(23-1)}{22}$

$=\dfrac{23^8\times 22}{22}$

$=23^8$

$RHS=23^x$

On comparing

$LHS$ and

$RHS$ we get,

$x=8$

Simplify :

$\displaystyle \frac{2^{2009}-2^{2007}}{2^{2006}-2^{2008}}$

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

Explanation

$\displaystyle \frac{2^{2009}-2^{2007}}{2^{2006}-2^{2008}}=\frac{2^{2007}(2^2-1)}{-(2)^{2006}(2^2-1)}$

$=-2^{2007-2006}$

$=-2$
Option B is correct.

The value of

$5^{1/4}\, \times\, (125)^{0.25}$ is

0%
$\sqrt 5$
0%
$5$
0%
$5\sqrt 5$
0%
$25$

Explanation

$5^{1/4} \times (125)^{0.25} = 5^{0.25} \times (5^3)^{0.25}$

$= 5^{0.25} \times 5^{(3\times 0.25)}$

$= 5^{0.25} \times 5^{0.75}$

$= 5^{(0.25 + 0.75)} = 5^1 = 5$

Simplify:

$\displaystyle \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$

0%
$5^5$
0%
$5^4$
0%
$5^2$
0%
$1$

Explanation

$\cfrac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$

$=\cfrac { 3^{ -5 }\times (2\times 5)^{ -5 }\times 5^{ 3 } }{ 5^{ -7 }\times (2\times 3)^{ -5 } }$

$=\cfrac { 3^{ -5 }\times 2^{ -5 }\times 5^{ -5 }\times 5^{ 3 } }{ 5^{ -7 }\times 2^{ -5 }\times 3^{ -5 } }$

$=3^{ -5+5 }\times 2^{ -5+5 }\times 5^{ -5+3+7 }$

$=5^{5}$

$=3125$

$\displaystyle \left ( \frac{-4}{5} \right )^{4} \times \left ( \frac{-4}{5} \right )^{2} = \frac{16}{25}$

0%
True
0%
False
0%
Ambiguous
0%
Data insffucient

Explanation

$(-\frac { 4 }{ 5 } )^{ 4 }*(-\frac { 4 }{ 5 } )^{ 2 }=(-\frac { 4 }{ 5 } )^{ 4+2 }=(-\frac { 4 }{ 5 } )^{ 6 }\neq \frac { 16 }{ 25 }$

$\displaystyle \left \{ \left ( \frac{5}{3} \right )^{15} \right \}^0$ is equal to

0%
$\displaystyle \frac{5}{3}$
0%
$\displaystyle \frac{3}{5}$
0%
1
0%
None of these

State true or false:

$\displaystyle \left ( \frac{2}{3} \right )^{4} \div \left ( \frac{2}{3} \right )^{6} = \left ( \frac{2}{3} \right )^{2}$

0%
True
0%
False
0%
Ambiuous
0%
Data insufficient

Explanation

$\left(\cfrac { 2 }{ 3 } \right)^{ 4 }\div \left(\cfrac { 2 }{ 3 } \right)^{ 6 }=\left(\cfrac { 2 }{ 3 } \right)^{ 4 } \times \left(\cfrac { 2 }{ 3 } \right)^{ -6 }=\left(\cfrac { 2 }{ 3 } \right)^{ 4-6 }=\left(\cfrac { 2 }{ 3 } \right)^{ -2 }$

State true or false:

$\displaystyle \left [ \left ( \frac{3}{7} \right )^{2}\right ]^3 = \left ( \frac{3}{7} \right )^{5}$

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

$\left[\left(\cfrac { 3 }{ 7 } \right)^{ 2 }\right]^{ 3 }=\left(\cfrac { 3 }{ 7 } \right)^{ 2 \times 3 }=\left(\cfrac { 3 }{ 7 } \right)^{ 6 }$

State true or false:

$\displaystyle \left ( \frac{-7}{9} \right )^{2} \div \frac{49}{81} = - 1$

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

$\left(\cfrac { -7 }{ 9 } \right)^{ 2 }\div \cfrac { 49 }{ 81 } =\left(\cfrac { -1 \times 7 }{ 9 } \right)^{ 2 }\times \left(\cfrac { 7 }{ 9 } \right)^{ -2 }=\left(\cfrac { 7 }{ 9 } \right)^{ 2-2 }=1$

Here,

$\left(\cfrac { -7 }{ 9 } \right)^{ 2 }=\left(\cfrac { 7 }{ 9 } \right)^{ 2 }$

(As the square of

$-ve$ sign number, will give a positive number)

Which expression is equivalent to 81?

0%
$2^9$
0%
$\displaystyle \left ( \frac{1}{3} \right )^{-4}$
0%
$3^{-4}$
0%
$\displaystyle \left ( \frac{1}{3} \right )^{4}$

Explanation

$({ \frac { 1 }{ 3 } ) }^{ -4 }\quad =\quad 3^{ 4 }\quad =\quad 81$

$\displaystyle \left ( \frac{1}{3} \right )^{7-7} = 2^0$

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

The value of

$\displaystyle \left ( \frac{5}{3} \right )^{-8} \div \left ( \frac{5}{3} \right )^{-8}$ is equal to

0%
$\displaystyle \frac{5}{3}$
0%
$\displaystyle \frac{3}{5}$
0%
$1$
0%
None of these

Explanation

According to the properties of exponents,

$\dfrac{a^m}{a^n}=a^{m-n}$

According to the given condition,

$\left(\cfrac { 5 }{ 3 } \right)^{ -8 }\div \left(\cfrac { 5 }{ 3 } \right)^{ -8 }=\left(\cfrac { 5 }{ 3 } \right)^{ -8-(-8) } \\= \left(\cfrac { 5 }{ 3 } \right)^{ 0 } \\=1$

Simplification of

$\displaystyle \left ( \frac{3}{5} \right )^{3} \times \left ( \frac{15}{2} \right )^{3}$ is

$\dfrac{729}{8}$

0%
True
0%
False

Explanation

$\displaystyle \left ( \frac{3}{5} \right )^{3} \times \left ( \frac{15}{2} \right )^{3}$

$=\displaystyle \left ( \frac{3}{5} \right )^{3} \times \left ( \frac{3\times 5}{2} \right )^{3}$

$=\displaystyle \left ( \frac{3^3}{5^3} \right ) \times \left ( \frac{3^3\times 5^3}{2^3} \right )$

$=\displaystyle \left ( \frac{3^3}1 \right ) \times \left ( \frac{3^3}{2^3} \right )$

$=\dfrac{3^6}{2^3}$

$=\dfrac{729}{8}$

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 2 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 2 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.