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

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 and comment at the end of the page.
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.

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

=   1/23x2
=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. $\left \{ \left (\frac {-2}{3 }\right )^{-2} \right \}^2$
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. $(5^{-1} \times 2^{-1}) \times 6^{-1}$
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 5m   ÷  5-3 = 55

5m   ÷   5-3 = 55
5m × (1/5-3)  = 55
5m+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. $\left ( \frac {5}{8} \right )^{-7} \times \left ( \frac {8}{5} \right )^{-4}}$
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.