In this page we have NCERT book Solutions for Class 8th Maths Chapter 12:Exponents and Power for
Exercise 12.1 . Hope you like them and do not forget to like , social share
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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
=1/2
6

(iv )

= (3
-7/ 3
-10) × 3
-5
=3
-7+10 × 3
-5
=3
3 × 3
-5
=3
-2
=1/3
2
(v) 2
- 3 × ( - 7)
- 3
= (2× -7)
-3
= (-14)
-3
=1/ (-14)
3
Question 3
Find the value of
- (30 + 4 - 1) × 22
- (2 - 1 × 4 - 1) ÷ 2 - 2
- $(\frac {1}{2})^{-2} + (\frac {1}{3})^{-2} + (\frac {1}{4})^{-2}$
- (3 - 1 + 4 - 1 + 5 - 1)0
- $\left \{ \left (\frac {-2}{3 }\right )^{-2} \right \}^2$
Answer
1.(3
0 + 4
- 1) × 2
2
= (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.
$(\frac {1}{2})^{-2} + (\frac {1}{3})^{-2} + (\frac {1}{4})^{-2}$
= 2
2 + 3
2 + 4
2
=4+9+16=29
4. (3
- 1 + 4
- 1 + 5
- 1)
0
=1 as a
0=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 = 5
5
Answer
5
m ÷ 5
-3 = 5
5
5
m × (1/5
-3) = 5
5
5
m+3 =5
5
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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