Find x such that 3 5 3 × 3 5 - 6 = 3 5 2 x - 1 .

3 5 3 × 3 5 - 6 = 3 5 2 x - 1 ⇒ 3 5 3 + ( - 6 ) = 3 5 2 x - 1 s ince a m × a n = a m + n ⇒ 3 5 - 3 = 3 5 2 x - 1 On equating the exponents: − 3 = 2 x − 1 ⇒ 2 x = − 3 + 1 ⇒ 2 x = − 2 &there4; x = - 2 2 = - 1

Simplify: 3 5 × 10 5 × 25 5 7 × 6 5 .

3 5 × 10 5 × 25 5 7 × 6 5 = 3 5 × 2 × 5 5 × 5 2 5 7 × 2 × 3 5 = 3 5 × 2 5 × 5 5 × 5 2 5 7 × 2 5 × 3 5 = 3 5 × 2 5 × 5 7 3 5 × 2 5 × 5 7 = 3 5 - 5 × 2 5 - 5 × 5 7 - 7 = 3 0 × 2 0 × 5 0 = 1 × 1 × 1 = 1

Simplify: 16 × 2 n + 1 - 4 × 2 n 16 × 2 n + 2 - 2 × 2 n + 2 .

The given expression is 16 × 2 n + 1 - 4 × 2 n 16 × 2 n + 2 - 2 × 2 n + 2 . The given expression can be written as 2 4 × 2 n + 1 - 2 2 × 2 n 2 4 × 2 n + 2 - 2 n + 1 × 2 2 = 2 2 × 2 2 × 2 n + 1 - 2 2 × 2 n 2 2 × 2 2 × 2 n + 2 - 2 n + 1 × 2 2 When multiplying like bases, keep the base the same and add the exponents. We know that, a m × a n = a m + n . Therefore, = 2 2 × 2 n + 1 + 2 - 2 2 × 2 n 2 2 × 2 n + 2 + 2 - 2 n + 1 × 2 2 = 2 2 × 2 n + 3 - 2 n 2 2 × 2 n + 4 - 2 n + 1 ⇒ 2 2 × 2 n + 3 - 2 n 2 2 × 2 n + 4 - 2 n + 1 = 2 n × 2 3 - 2 n 2 n × 2 4 - 2 n × 2 1 = 2 n 2 3 - 1 2 n 2 4 - 2 = 8 - 1 16 - 2 = 7 14 = 1 2 Hence, value of the expression 16 × 2 n + 1 - 4 × 2 n 16 × 2 n + 2 - 2 × 2 n + 2 is 1 2 .

Find the value of n when: 5 2 n × 5 3 = 5 9 , 8 × 2 n + 2 = 32 and 6 2 n + 1 ÷ 36 = 6 3

(i) 5 2n × 5 3 = 5 9 5 2n +3 = 5 9 [since a n × a m = a m+n ] On equating the coefficients: 2n + 3 = 9 ⇒ 2n = 9 − 3 ⇒ 2n = 6 &there4; n = 6 2 = 3 (ii) 8 × 2 n +2 = 32 ⇒ (2) 3 × 2 n +2 = (2) 5 [since 2 3 = 8 and 2 5 = 32] ⇒ (2) 3+ (n+2) = (2) 5 On equating the coefficients: 3 + n + 2 = 5 ⇒ n + 5 = 5 ⇒ n = 5 − 5 &there4; n = 0 (iii) 6 2n +1 ÷ 36 = 6 3 ⇒ 6 2n +1 ÷ 6 2 = 6 3 [since 36 = 6 2 ] ⇒ 6 2 n + 1 6 2 = 6 3 ⇒ 6 2 n + 1 - 2 = 6 3 [since a m a n = a m - n ] ⇒ 6 2n-1 = 6 3 On equating the coefficients: 2n - 1 = 3 ⇒ 2n = 3 + 1 ⇒ 2n = 4 &there4; n = 4 2 = 2

If 2 n − 7 × 5 n − 4 = 1250 , find the value of n.

2 n - 7 × 5 n - 4 = 1250 ⇒ 2 n 2 7 × 5 n 5 4 = 2 × 5 4 [ sin c e 1250 = 2 × 5 4 ] ⇒ 2 n × 5 n 2 7 × 5 4 = 2 × 5 4 ⇒ 2 n × 5 n = 2 × 5 4 × 2 7 × 5 4 [using cross multiplication] ⇒ 2 n × 5 n = 2 1 + 7 × 5 4 + 4 [ sin c e a m × a n = a m + n ] ⇒ 2 n × 5 n = 2 8 × 5 8 ⇒ 2 × 5 n = 2 × 5 8 [ sin c e a n × b n = ( a × b ) n ] ⇒ 10 n = 10 8 ⇒ n = 8

E M B I B E

MATHEMATICS FOR CLASS 7>Chapter 5 - Exponents>EXERCISE 5A

R S Aggarwal Solutions for Chapter: Exponents, Exercise 1: EXERCISE 5A

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Chapter: Exponents, Exercise 1: EXERCISE 5A

Attempt the free practice questions on Chapter 5: Exponents, Exercise 1: EXERCISE 5A with hints and solutions to strengthen your understanding. MATHEMATICS FOR CLASS 7 solutions are prepared by Experienced Embibe Experts.