NCERT Solutions Class 8 - Chapter 10 Exponents and Powers – Exercise 10.1

Exponents and Powers are ways to represent very large numbers or small numbers in a simplified format. Basically Exponents and Power is an expression for repeated multiplications of the same number or factor. This article covers some problems and laws.

Laws of Exponents

1. Multiplication Law
As per the multiplication law of exponents, the product of two exponents with the same base and different powers equals to base raised to the sum of the two powers or integers.

a^m × aⁿ = a^{m+n}

2. Division Law

When two exponents having the same bases and different powers are divided, then it results in the base being raised to the difference between the two powers.

a^m ÷ a^n = a^m / a^n = a^{m-n}

3. Negative Exponent Law

Any base if has a negative power, then it results in reciprocal but with positive power or integer to the base.

a^{-m} = 1/a^m

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 = aⁿ if 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

Summary

This article covers essential things that should be known for basic exponents and power problems. Exponent is expression of multiplication. Refer this article for more practice problems.