Question

If $\frac{9^n\times 3^2\times {\left(3^{{\displaystyle \frac{-n}{2}}}\right)}^{-2}-{\left(27\right)}^n}{3^{3m}\times 2^3}=\frac{1}{27}$, prove that m – n = 1.

Answer 1

$$\begin{gathered} \frac{9^n\times 3^2\times {\left(3^{{\displaystyle \frac{-n}{2}}}\right)}^{-2}-{\left(27\right)}^n}{3^{3m}\times 2^3}=\frac{1}{27} \\ \Rightarrow \frac{{\left(3^2\right)}^n\times 3^2\times {\left(3^{{\displaystyle -n}}\right)}^{-1}-{\left(3^3\right)}^n}{3^{3m}\times 2^3}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{2n}\times 3^2\times 3^{{\displaystyle n}}-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{2n+2+n}-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{3n+2}-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{3n}\times 3^2-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{3n}\left(9-1\right)}{3^{3m}\times 8}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{3n}\left(8\right)}{3^{3m}\times 8}=\frac{1}{3^3} \\ \Rightarrow \frac{3^{3n}}{3^{3m}}=\frac{1}{3^3} \\ \Rightarrow 3^{3n-3m}=3^{-3} \\ \Rightarrow 3n-3m=-3 \\ \Rightarrow 3\left(n-m\right)=-3 \\ \Rightarrow n-m=-1 \\ \Rightarrow m-n=1 \end{gathered}$$
Hence, m – n = 1.