Question

Find the value of ’n’ in each of the following:

(i) (2/3)³ × (2/3)⁵ = (2/3)ⁿ⁻²

(ii) (-3)ⁿ⁺¹ x (-3)⁵ = (-3)³

(iii) 7²ⁿ⁺¹ ÷ 49 = 7³

Answer 1

(i) (2/3)³ × (2/3)⁵ = (2/3)ⁿ⁻²

Here bases are equal, so exponents are also equal.

⇒ n – 2 = 8 

⇒ n = 8 + 2 = 10 

∴ n = 10

(ii) (-3)ⁿ⁺¹ x (-3)⁵ = (-3)³

⇒ (-3)ⁿ⁺¹⁺⁵ = (-3)^-4  [∵ a^m x aⁿ = a^m+n ]

⇒ (-3)ⁿ⁺⁶ = (-3)^-4

⇒ n + 6 = -4

⇒ n = -4 – 6 = -10

⇒ n = -10

(iii) 7²ⁿ⁺¹ ÷ 49 = 7³    

⇒ 7^2n+1-2 = 7³ [ ∵ a^m/aⁿ = a^m−n ]

⇒ 7^{2n – 1}= 7³

⇒ 2n – 1 = 3

⇒ 2n = 3 + 1 = 4

⇒ n = 4/2

∴ n = 2