Question

9^n×3^2×3^n-(27)^n÷(3^3)^5×2^3 = 1/27​

Answer 1

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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