Question

can anyone help me with this question​

Answer 1

Given

(9n × 32 × 3n – (27)n)/ (33)5 × 23 =

(1/27) =

(32)n × 33 × 3n – (33)n/ (315 × 23) =

(1/27) = 3(2n+2+n) – (33)n/ (315 × 23)

= (1/27)

= 3(3n+2)– (33)n/ (315 × 23) =

(1/27) =

33n × 32 – 33n/ (315 × 23) =

(1/27) =

33n × (32 – 1)/ (315 × 23) =

(1/27) =

33n × (9 – 1)/ (315 × 23) =

(1/27) =

33n × (8)/ (315 × 23) =

(1/27) =

33n × 23/ (315 × 23) =

(1/27) =

33n/315 =

(1/27) = 33n-15 =

(1/27) = 33n-15 =

(1/33) = 33n-15 = 3-3

On equating the coefficients, we get

3n -15 = -3 ⇒ 3n = -3 + 15 ⇒ 3n = 12 ⇒ n = 12/3 = 4

Answer 2

Answer:

(32)n × 33 × 3n – (33)n/ (315 × 23) = (1/27) = 3(2n+2+n) – (33)n/ (315 × 23) = (1/27) = 3(3n+2)– (33)n/ (315 × 23) = (1/27) = 33n × 32 – 33n/ (315 × 23) = (1/27) = 33n × (32 – 1)/ (315 × 23) = (1/27) = 33n × (9 – 1)/ (315 × 23) = (1/27) = 33n × (8)/ (315 × 23) = (1/27) = 33n × 23/ (315 × 23) = (1/27) = 33n/315 = (1/27) = 33n-15 = (1/27) = 33n-15 = (1/33) = 33n-15 = 3-3 On equating the coefficients, we get 3n -15 = -3 ⇒ 3n = -3 + 15 ⇒ 3n = 12 ⇒ n = 12/3 = 4Read more on Sarthaks.com - https://www.sarthaks.com/682795/if-9-n-3-2-3-n-27-n-3-3-5-2-3-1-27-find-the-value-of-n?show=682798#a682798