NCERT Solutions for Class 8 Maths Chapter 12 Exponents CBSE Exercise 12.1

Question 1
Evaluate
(i) $3^{-2}$
(ii) $(-4)^{-2}$
(iii)$(\frac {1}{2})^{-5}$ Answer
As we know that
$b^{-n}= \frac {1}{b^n}$
(i) $3^{-2}=\frac {1}{3^2} =\frac {1}{9}$
(ii) $(-4)^{-2}=\frac {1}{(-4)^2}= \frac {1}{16}$
(iii) $(\frac {1}{2})^{-5}=\frac {1}{2^{-5}}=2^5= 32$
Question 2
Simplify and express the result in power notation with positive exponent.

Answer
(i)
= (-4)⁵ / (-4)⁸
= (-4)^5-8
=1/ (-4)³ (ii)

=   1/2^3x2
=1/2⁶

(iv )

=   (3^-7/ 3^-10) × 3^-5
=3^-7+10 × 3^-5
=3³ × 3^-5
=3^-2
=1/3²
(v) 2 ^{- 3} × ( - 7) ^{- 3}
=   (2× -7)^-3
= (-14)^-3
=1/ (-14)³
Question 3
Find the value of

1. (3⁰ + 4 ^{- 1}) × 2²
2. (2 ^{- 1} × 4 ^{- 1})  ÷ 2 ^{- 2}
3. $(\frac {1}{2})^{-2} + (\frac {1}{3})^{-2} + (\frac {1}{4})^{-2}$
4. (3 ^{- 1} + 4 ^{- 1} + 5 ^{- 1})⁰
5. $\left \{ \left (\frac {-2}{3 }\right )^{-2} \right \}^2$

Answer
1.(3⁰ + 4 ^{- 1}) × 2² = (1+1/4) × 4
= 4+1=5
2. (2 ^{- 1} × 4 ^{- 1})  ÷ 2 ^{- 2}
= [(1/2) × (1/4)] ÷ (1/4)
= (1/8)  ×  (4)  =1/2
3.
$(\frac {1}{2})^{-2} + (\frac {1}{3})^{-2} + (\frac {1}{4})^{-2}$ = 2² + 3²  + 4²
=4+9+16=29
4. (3 ^{- 1} + 4 ^{- 1} + 5 ^{- 1})⁰
=1   as  a⁰=1
5.$\left \{ \left (\frac {-2}{3 }\right )^{-2} \right \}^2$
$=\left (\frac {-2}{3 }\right )^{-4}$
$=\frac {81}{16}$
Question 4
Evaluate
i. $ \frac {8^{-1} \times 5^3}{2^{-4}}$
ii. $(5^{-1} \times 2^{-1}) \times 6^{-1}$
Answer
i. $ \frac {8^{-1} \times 5^3}{2^{-4}}$
$=\frac {2^4 \times 5^3}{8} = 250$
ii. $(5^{-1} \times 2^{-1}) \times 6^{-1}$
$=( \frac {1}{5} \times \frac {1}{2} ) \times \frac {1}{6}$
$=\frac {1}{60}$


Question 5
Find the value of m for which 5^m   ÷  5^-3 = 5⁵

Answer
5^m   ÷   5^-3 = 5⁵
5^m × (1/5^-3)  = 5⁵
5^m+3 =5⁵
So  m+3=5
m=2
Question 6
Evaluate
i. $\left \{ \left ( \frac {1}{3} \right )^{-1} - \left ( \frac {1}{4} \right )^{-1} \right \}^{-1}$
ii. $\left ( \frac {5}{8} \right )^{-7} \times \left ( \frac {8}{5} \right )^{-4}}$
Answer
i. $\left \{ \left ( \frac {1}{3} \right )^{-1} - \left ( \frac {1}{4} \right )^{-1} \right \}^{-1}$
$={3^2 - 4^1}^{-1} = -1$ ii. $\left ( \frac {5}{8} \right )^{-7} \times \left ( \frac {8}{5} \right )^{-4}}$
$=\left ( \frac {5}{8} \right )^{-7} \times \left ( \frac {5}{8} \right )^{4}}$
$=\left ( \frac {5}{8} \right )^{-3}= \frac {512}{125}$
Question 7
Simplify.


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