GSEB Solutions Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1

Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1

Question 1.
Evaluate:
(i) 3^-2
(ii) (-4)^-2
(iii) (\(\frac{1}{2}\))^-5
Solution:

Question 2.
Simplify and express the result in power notation with positive exponent?
(i) (-4)⁵ + (-4)⁸
(ii) \(\left(\frac{1}{2^{3}}\right)^{2}\)
(iii) (-3)⁴ × (\(\frac{5}{3}\))⁴
(iv) (3^-7 + 3^-10) × 3^-5
(v) 2^-3 × (-7)^-3
Solution:
(i) (-4)⁵ + (-4)⁸
∵ a^m + aⁿ = a^m-n
∴ (-4)⁵ + (-4)⁸ = (-4)^5-8
= (-4)^-3 = \(\frac{1}{(-4)^{3}}\)

(ii)

(iii)

= [(-1) × 5]⁴ = [(-1)⁴ × (+5)⁴]
= 1 × (5)⁴ = (5)⁴

(iv) (3^-7 ÷ 3^-10) × 3^-5
∵ a^m + aⁿ = a^m-n and a^m × aⁿ = a^m+n
∴ (3^-7 ÷ 3^-10 × 3^-5 = [3^-7-(-10)] × 3^-5
= 3^3+(-5) = 3^-2 = \(\frac{1}{(3)^{2}}\)

(v) 2^-3 × (-7)^-3
∵ a^m × b^m = (ab)^m
∴ 2^-3 × (-7)^-3 = [2 × (-7)]^-3
= [-14]^-3 = \(\frac{1}{(-14)^{3}}\)

Question 3.
Find the value of:
(i) (3⁰ + 4^-1) × 2²
(ii) (2^-1 × 4^-1) + 2^-2
(iii) (\(\frac{1}{2}\))^-2 + (\(\frac{1}{3}\))^-2 + (\(\frac{1}{4}\))^-2
(iv) (3^-1 + 4^-1 + 5^-1)⁰
(v) {(\(\frac{-2}{3}\))^-2}²
Solution:
(i) (3⁰ + 4^-1) × 2²
∵ a⁰ = 1 and a^-1 = \(\frac{1}{a}\)
∴ (3⁰ + 4^-1) × 2²
= (1 + \(\frac{1}{2}\)) 2² = \(\frac{5}{4}\) × 4 = 5

(ii) (2^-1 × 4^-1) ÷ 2^-2
∵ a^m × b^m = (ab)^m
∴ (2^-1 × 4^-1) + 2^-2 = (2 × 4^-1) ÷ 2^-2
= (2¹ × 2²)^-1 ÷ 2^-2
= (2¹⁺²)^-1) ÷ 2^-2 = (2³)^-1 ÷ 2^-2
= 2^-3 ÷ 2^-2 = (2)^(-3)-(-2)
= (-2)^-3+2 = 2^-1 = \(\frac{1}{2}\)

(iii)

(iv)

(v)

Question 4.
Evaluate

1. \(\frac{8^{-1} \times 5^{3}}{2^{-4}}\) and
2. (5^-1 × 2^-1) × 6^-1

Solution:
1. \(\frac{8^{-1} \times 5^{3}}{2^{-4}}\) = \(\frac{8^{-1} \times(5 \times 5 \times 5)}{2^{-4}}\)
= \(\frac{1}{8}\) × 2⁴ × 125 = \(\frac{1}{8}\) × 2 × 2 × 2 × 2 × 125
= 2 × 125 = 250

2. (5^-1 × 2^-1) × 6^-1) = (5 × 2^-1) × 6^-1)
= 10^-1 × 6^-1
= (10 × 6^-1) = 60^-1 = \(\frac{1}{60}\)

Question 5.
Find the value of m for which 5^m + 5^-3 = 5⁵.
Solution:
∵ a^m × b^m = (ab)^m
∴ L.H.S = 5^m + 5^-3
= 5^m-(-3)
= 5^m+3
Now, 5^m+3 = 5⁵
Since, bases 5 are equal, therefore exponents are also equal.
i.e; m + 3 = 5
or m = 5 – 3 = 2
Thus, the required value of m is 2.

Question 6.
Evaluate:
(i) \(\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
(ii) \(\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}\)
Solution:

Question 7.
Simplify:
(i) \(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}\) (t ≠ 0)
(ii) \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)
Solution: