Class 8 NCERT Solutions- Chapter 12 Exponents and Powers – Exercise 12.1

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Question 1. Evaluate:

Solution:

(i) 3^–2

3^-2 = $\frac{1}{(3^2)} = \frac{1}{9}$    (Property used: a^-n = $\frac{1}{a^n}$ )

(ii) (– 4)^{– 2}

(-4)^-2 = $\frac{1}{(-4)^2} = \frac{1}{16}$   (Property used: a^-n = $\frac{1}{a^n}$ )

(iii) ( $\frac{1}{2}$ ) ^-5

( $\frac{1}{2}$ )^-5 = (2)⁵= 32   (Property used: $(\frac{b}{a})^{-n} = \frac{a^n}{b^n}$ )

Question 2. Simplify and express the result in power notation with a positive exponent.

Solution:

(i) (-4)⁵ ÷ (-4)⁸

= (-4)^5-8= (-4) ^-3 (Property used: a^m ÷ aⁿ= a^m-n)

= $\frac{1}{(-4)^3}$

= $\mathbf{(\frac{1}{(-4)})^3}$

(ii) $(\frac{1}{2^3})^2$

= $\frac{(1)^2}{(2^3)^2}$ (Property used: (a^m)ⁿ = a^m×n)

= $\frac{1}{2^6}$

= $\mathbf{\frac{1}{2^6}}$

(iii) (-3)⁴ × ( $\frac{5}{3}$ )⁴

= ((3)⁴ × $\frac{5^4}{3^4}$ (Property used: (a/b)ⁿ = aⁿ/ bⁿ& (-a)ⁿ= aⁿif a is positive number and n is even)

= 5⁴

(iv) (3^-7 ÷ 3^-10) × 3^-5

= 3 ^(-7-(-10)) × 3^-5 (Property used: a^m ÷ aⁿ= a^m-n)

= 3 ^(-7+10) × 3^-5

= 3³ × 3^-5

= 3 ^(3+(-5))(Property used: a^m × aⁿ= a ^{m + n})

= 3^-2 (Property used: a^-m = $\frac{1}{a^m}$ )

= $\frac{1}{3^2}$

= $\mathbf{\frac{1}{3^2}}$

(v) 2^-3 × (-7)^-3

= (2 × (-7))^-3(Property used: a^m × b^m = (a×b)^m)

= (-14)^-3(Property used: a^-m = $\frac{1}{a^m}$ )

= $\frac{1}{(-14)^3}$

= $\mathbf{\frac{1}{(-14)^3}}$

Question 3. Find the value of

Solution:

(i) (3⁰+ 4^-1) × 2²

= (1 + ( $\frac{1}{4}$ )) × 4 (a⁰ = 1 a ≠ 0)

= ( $\frac{5}{4}$ ) × 4

= 5

(ii) (2^-1 × 4^-1) ÷ 2^-2

= (2 × 4)^-1 ÷ $\frac{1}{2^2}$ (Property used: a^m × b^m = (a×b)^m)

= (8)-1 ÷ $\frac{1}{4}$

= ( $\frac{1}{8}$ ) ÷ $\frac{1}{4}$

=( $\frac{1}{8}$ ) × 4

= ( $\frac{1}{2}$ )

(iii) (1/2)^-2 + (1/3)^-2 + (1/4)^-2

= 2²+ 3² + 4²(Property used: $(\frac{1}{a})^{-m}$ =a^m)

= 4 + 9 + 16

= 29

(iv) (3^-1 + 4^-1 + 5^-1)⁰

= ( $\frac{1}{3} + \frac{1}{4} + \frac{1}{5}$ )⁰(a⁰ = 1 (a ≠ 0)

= 0

(v) {( $\frac{-2}{3}$ )^-2}²

= ( $\frac{-2}{3}$ ) ^-2×2(Property used: (a^m)ⁿ= a^m×n)

= ( $\frac{-2}{3}$ )^-4 = ( $\frac{-3}{2}$ )⁴(Property used: (b/a)^-n = aⁿ/bⁿ)

= $\frac{3^4}{2^4}$

= $\frac{81}{16}$

Question 4. Evaluate

Solution:

(i) (8^-1 × 5³) / 2^-4

= ( $\frac{1}{8}$ × 125) / (2^-4) (Property used: (b/a)^-n = aⁿ/bⁿ)

= ( $\frac{1}{8}$ ) × 125 × 2⁴

= 250

(ii) (5^-1 × 2^-1) × 6^-1

= (5 × 2)^-1 × 6^-1 (Property used: a^m × b^m= (a×b)^m)

= 10^-1 × 6^-1

= (10 × 6)^-1 (Property used: a^m × b^m = (a×b)^m)

= 60^-1

= $\frac{1}{60}$

Question 5. Find the value of m for which 5^m ÷ 5^{– 3} = 5⁵

Solution:

5^{m-(– 3)} = 5⁵(Property used: a^m ÷ aⁿ= a^m-n)

5^m+3 = 5⁵

m+3 = 5

m = 5-3

m = 2

Question 6. Evaluate

Solution:

(i) {( $\frac{1}{3}$ )^-1 – ( $\frac{1}{4}$ )^-1}^-1

= (3¹ – 4¹) ^-1(Property used: (1/a)^-m = a^m)

= (-1)^-1

= (1/(-1))¹

= (-1)

(ii) ( $\frac{5}{8}$ )^-7 × ( $\frac{8}{5}$ )^-4

= ( $\frac{5}{8}$ )⁷ × ( $\frac{8}{5}$ )^-4(Property used: (b/a)^-n = (a/b)ⁿ)

= ( $\frac{8}{5}$ ) ^{7+ (-4)}(Property used: a^m× aⁿ = a ^{m + n})

= ( $\frac{8}{5}$ )³= 8³/5³

= $\frac{512}{125}$

Question 7. Simplify

Solution:

(i) $\frac{(25 \times t^{-4})}{(5^{-3} \times 10 \times t^{-8})}$ (t ≠ 0)

= $\frac{(5^2 \times t^{-4})}{(5^{-3} \times 10 \times t^{-8})}$ (Property used: a^m÷ aⁿ= a^m-n ) (25 = 5²)

= $\frac{(5^{2-(-3)} \times t^{-4-(-8)})}{ 10}$

= $\frac{(5^{5} \times t^{4})}{ 10}$

= $\frac{(625 \times t^{4})}{ 2}$

(ii) $\frac{(3^{-5} \times 10^{-5} \times 125)}{(5^{-7} \times 6^{-5})}$

= $\frac{(3^{-5} \times (2\times 5)^{-5} \times 125)}{(5^{-7} \times (2 \times 3)^{-5})}$

= $\frac{(3^{-5} \times 2^{-5}\times 5^{-5} \times 125)}{(5^{-7} \times 2^{-5} \times 3^{-5})}$ (Property used: (a×b)^m = a^m × b^m)

= (3^-5-(-5) × 2^-5-(-5) × 5 ^(-5)+3+7)(Property used: a^m ÷ aⁿ= a^m-n )

= (3⁰× 2⁰ × 5⁵)(a⁰= 1 (a ≠ 0)

= 5⁵

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Last Updated : 10 Mar, 2021