Question

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

Answer 1

$\frac{9^n\times 3^2\times {\left(3^{-n/2}\right)}^{-2}-{27}^n}{3^{3m}\times 2^3}=1/27\Rightarrow \frac{3^{2n}\times 3^2\times 3^n-3^n}{3^{3m}\times 2^3}=\frac{1}{3^3}\Rightarrow \frac{3^{3n}\times 3^2-3^{3n}}{3^{3m}\times 2^3}=\frac{1}{3^3}$

$\Rightarrow 3^{3n+2}-3^{3n}=\frac{3^{3m}\times 2^3}{3^3}$

$\Rightarrow 3^{3n}[3^2-1]=\frac{3^{3m}\times 2^3}{3^3}$

$\Rightarrow 3^{3n}\left[8\right]=\frac{3^{3m}\times 8}{3^3}$

$\Rightarrow 3^{3n}=3^{3m-3}$

$\Rightarrow$ Equating powers, we get

$3n=3m-3$

$\Rightarrow n=m-1$

$\Rightarrow n=m-1$

$\Rightarrow m-n=1$

Hence proved.