In this page we have Important questions for exponents & power Class 8 maths Chapter 12 CBSE
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Question 1
Evaluate
- 2^{-2}
- (-2)^{-2}
- (3/2)^{-5}
Solution
As we know that
b^{-n}= 1/b^{n}
(i) 2^{-2} = 1/ 2^{2} =1/4
(ii) (-2)^{-2} = 1/(-2)^{2}= 1/4
(iii) (3/2)^{-5 }=3-^{5}/ 2^{-5}
=2^{5}/3^{5 }= 32/243
Question 2
Simplify and express the result in power notation with positive exponent.
(i) (-2)
^{5} ÷ (-2)
^{4}
(ii) (1/2)
^{2} × (2/5)
^{2}
(iii) (-5)
^{2} × (3/5)
Solution
(i) (-2)^{5} ÷ (-2)^{4}
= (-2)^{5} / (-2)^{4}
= (-2)^{5-4}
=-2
(ii) (1/2)^{2} × (2/5)^{2}
= (1/4) X (4/25)
= 1/25
(iii) (-5)^{2} × (3/5)
=25 × (3/5)
=15
Question 3
Find the value of.
(i)(4
^{0} + 4
^{-1}) × 2
^{2}
(ii)(3
^{-1} × 9
^{-1}) ÷ 3
^{-2}
(iii)(11
^{-1} + 12
^{-1} + 13
^{-1})
^{0}
Solution
(i)(4^{0} + 4^{-1}) × 2^{2}
= (1+1/4) × 4
= 4+1=5
(ii)(3^{-1} × 9^{-1}) ÷ 3^{-2}
= [(1/3) × (1/9)] ÷ (1/9)
=1/3
(iii)(11^{-1} + 12^{-1} + 13^{-1})^{0}
=1 as a^{0}=1
Question 4
Find the value of x here
$( \frac {11}{9})^3 \times (\frac {9}{11})^6 = (\frac {11}{9})^{2x-1}$
Solution
$( \frac {11}{9})^3 \times (\frac {9}{11})^6 = (\frac {11}{9})^{2x-1}$
$(\frac {11}{9})^{3-6} = (\frac {11}{9})^{2x-1}$
-3=2x-1
Or
x=-1
Question 5
Find the value of m for which 2
^{m} ÷ 2
^{-4} = 4
^{5}
Solution
2^{m} ÷ 2^{-4} = 4^{5}
2^{m} × (1/2^{-4}) = 2^{10}
2^{m+4} =2^{10}
So m+4=10
m=6
Question 6
Express the following numbers in standard form.
(i) 0.0000000015
(ii) 0.00000001425
(iii) 102000000000000000
Solution
(i)0.0000000015
=1.5 ×10^{-9}
(ii)0.00000001425
=1.425×10^{-8}
(iii)102000000000000000
=1.02×10^{17}
Question 7
Express the following numbers in usual form.
(i) 34.02 x 10
^{-5}
(ii) 9.5 x 10
^{5}
(iii) 9 x 10
^{-4}
(iv) 2.0001 x 10
^{8}
Solution
(i)34.02 x 10^{-5}
= .0003402
(ii)9.5 x 10^{5}
=950000
(iii) 9 x 10^{-4}
=.0009
(iv)2.0001 x 10^{8}
=200010000
Question 8
Solve for the variables
(i) $\frac {1}{y^{-5}} = 32$
(ii) $ (\frac {1}{7})^3 \div (\frac {1}{7})^8 = 7^n$
(iii) $ \frac {3^x .3}{3^{2x+1}} = 27$
Solution
(i) $\frac {1}{y^{-5}} = 32$
$y^5 = 32$
$y^5 = 2^5$
Therefore y=2
(ii) $ (\frac {1}{7})^3 \div (\frac {1}{7})^8 = 7^n$
$(\frac {1}{7})^3 \times 7^8 = 7^n$
$7^5 = 7^n$
Hence n=5
(iii) $ \frac {3^x .3}{3^{2x+1}} = 27$
$\frac {3^{x+1}}{3^{2x+1}} = 27$
$3^{x + 1 - 2x -1} =27$
$3^{-x}= 27$
$3^{-x}=3^3$
Hence x =-3
Question 9
Find the Multiplicative inverse of
(i) $3^{25}$
(ii) $4^{-3}$
(iii) $(\frac {2}{3})^{-2}$
Solution
(i) $3^{-25}$
(ii) $4^{3}$
(iii) $(\frac {2}{3})^{2}$
Question 10
Simplify the following
(i) $1 + \left [ \left ( \frac{1}{3} \right )^{-3} - \left ( \frac{1}{2} \right )^{-3} \right ] \div 38$
(ii) $\frac {1}{1 + p^{a-b}} + \frac {1}{1 + p^{b-a}}$
Solution
(i) $1 + \left [ \left ( \frac{1}{3} \right )^{-3} - \left ( \frac{1}{2} \right )^{-3} \right ] \div 38$
=$1 + (3^3 - 2^3) \div 38$
=$1 + 19 \div 38$
=$\frac {3}{2}$
(ii) $\frac {1}{1 + p^{a-b}} + \frac {1}{1 + p^{b-a}}$
=$\frac {p^b}{p^b + p^a} + \frac {p^a}{p^a + p^b}$
=$\frac {p^b + p^a}{p^b + p^a}$
=1
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