Solve for n:

$\dfrac{3^n \times 9^{n + 1}}{3^{n - 1} \times 27^{n - 1}} = 81$

Given,

$\dfrac{3^n \times 9^{n + 1}}{3^{n - 1} \times 27^{n - 1}} = 81$

Solving for n,

$\Rightarrow \dfrac{3^n \times 9^{n + 1}}{3^{n - 1} \times 27^{n-1}} = 81 \\[1em] \Rightarrow \dfrac{3^n \times [(3)^2]^{n + 1}}{3^{n - 1} \times [(3)^3]^{n-1}} = 3^4 \\[1em] \Rightarrow \dfrac{3^n \times 3^{2n + 2}}{3^{n - 1} \times (3)^{3n - 3}} = 3^4 \\[1em] \Rightarrow \dfrac{3^{2n + 2 + n}}{3^{n - 1 + 3n - 3}} = 3^4 \\[1em] \Rightarrow \dfrac{3^{3n + 2}}{3^{4n - 4}} = 3^4 \\[1em] \Rightarrow 3^{3n + 2 - (4n - 4)}= 3^4 \\[1em] \Rightarrow 3^{3n + 2 - 4n + 4}= 3^4 \\[1em] \Rightarrow 3^{6 - n}= 3^4$

Equating the exponents,

⇒ 6 - n = 4

⇒ n = 6 - 4 = 2.

Hence, n = 2.