MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 2 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Logarithm And Antilogarithm
Quiz 2
The value of $$(\sqrt 8)^{\tfrac 13}$$ is
Report Question
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 $$(32)^{\displaystyle \frac {-2}{5}}\, \div\, (125)^{\displaystyle \frac {-2}{3}}$$
Report Question
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 $$\displaystyle \frac {1}{x^y}\, +\, \displaystyle \frac {1}{y^x}$$ = .........
Report Question
0%
72
0%
$$\displaystyle \frac {17}{72}$$
0%
$$\displaystyle \frac {31}{108}$$
0%
None of these
Explanation
$$\displaystyle \frac {1}{2^3}\, +\, \displaystyle \frac {1}{3^2}\, =\, \displaystyle \frac {1}{8}\, +\, \displaystyle \frac {1}{9}\, =\, \displaystyle \frac {9+8}{72}\, =\, \displaystyle \frac {17}{72}$$
$$\left ( \displaystyle \frac {16}{81} \right )^{\displaystyle \frac {3}{4}}$$ = ..........
Report Question
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
Report Question
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
Report Question
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
Report Question
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 $$2^{0.64}*2^{0.36}$$ ?
Report Question
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
Report Question
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. $$4^1$$
Report Question
0%
I and II
0%
II and III
0%
I and III
0%
I and IV
Explanation
I $$1^4 = 1$$
II: $$4^0 = 1$$
III: $$0^4 = 0$$
IV: $$4^1 = 4$$
Hence, $$1^4=4^0 = 1$$
Evaluate: $$(256)^{0.16}\, \times\, (256)^{0.09}$$
Report Question
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 $$[\log_{10} (5\log_{10} 100)]^{2}$$?
Report Question
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:
Report Question
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
Report Question
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
Report Question
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: $$(0.04)^{-1.5} $$
Report Question
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 :
Report Question
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.
if $$\dfrac{(23)^{9}-(23)^{8}}{22}=(23)^{x}$$ then $$x$$ equals :
Report Question
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 : $$\displaystyle \frac{2^{2009}-2^{2007}}{2^{2006}-2^{2008}}$$
Report Question
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
Report Question
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: $$\displaystyle \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$$
Report Question
0%
$$5^5$$
0%
$$5^4$$
0%
$$5^2$$
0%
$$1$$
Explanation
$$ \cfrac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$$
$$=\cfrac { 3^{ -5 }\times (2\times 5)^{ -5 }\times 5^{ 3 } }{ 5^{ -7 }\times (2\times 3)^{ -5 } } $$
$$=\cfrac { 3^{ -5 }\times 2^{ -5 }\times 5^{ -5 }\times 5^{ 3 } }{ 5^{ -7 }\times 2^{ -5 }\times 3^{ -5 } } $$
$$=3^{ -5+5 }\times 2^{ -5+5 }\times 5^{ -5+3+7 }$$
$$=5^{5}$$
$$=3125$$
$$\displaystyle \left ( \frac{-4}{5} \right )^{4} \times \left ( \frac{-4}{5} \right )^{2} = \frac{16}{25}$$
Report Question
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
Report Question
0%
$$\displaystyle \frac{5}{3}$$
0%
$$\displaystyle \frac{3}{5}$$
0%
1
0%
None of these
Explanation
Anything raised to the power of 0 equals 1
State true or false:
$$\displaystyle \left ( \frac{2}{3} \right )^{4} \div \left ( \frac{2}{3} \right )^{6} = \left ( \frac{2}{3} \right )^{2}$$
Report Question
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:
$$\displaystyle \left [ \left ( \frac{3}{7} \right )^{2}\right ]^3 = \left ( \frac{3}{7} \right )^{5}$$
Report Question
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: $$\displaystyle \left ( \frac{-7}{9} \right )^{2} \div \frac{49}{81} = - 1$$
Report Question
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?
Report Question
0%
$$2^9$$
0%
$$\displaystyle \left ( \frac{1}{3} \right )^{-4}$$
0%
$$3^{-4}$$
0%
$$\displaystyle \left ( \frac{1}{3} \right )^{4}$$
Explanation
$$({ \frac { 1 }{ 3 } ) }^{ -4 }\quad =\quad 3^{ 4 }\quad =\quad 81$$
$$\displaystyle \left ( \frac{1}{3} \right )^{7-7} = 2^0$$
Report Question
0%
True
0%
False
0%
Ambiguous
0%
Data insufficient
Explanation
7-7 = 0,
anything raised to the power of 0 equals 1
The value of $$\displaystyle \left ( \frac{5}{3} \right )^{-8} \div \left ( \frac{5}{3} \right )^{-8}$$ is equal to
Report Question
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 $$\dfrac{729}{8}$$
Report Question
0%
True
0%
False
Explanation
$$\displaystyle \left ( \frac{3}{5} \right )^{3} \times \left ( \frac{15}{2} \right )^{3}$$
$$=\displaystyle \left ( \frac{3}{5} \right )^{3} \times \left ( \frac{3\times 5}{2} \right )^{3}$$
$$=\displaystyle \left ( \frac{3^3}{5^3} \right ) \times \left ( \frac{3^3\times 5^3}{2^3} \right )$$
$$=\displaystyle \left ( \frac{3^3}1 \right ) \times \left ( \frac{3^3}{2^3} \right )$$
$$=\dfrac{3^6}{2^3}$$
$$=\dfrac{729}{8}$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
