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

Upon a little simplification,

$\frac{(3^{2n}\times 3^2\times 3^n)-3^{3n}}{3^{3m}}$ = $\frac{8}{27}$

$\Rightarrow$$3^{3n-3m+2}-3^{3n-3m}$ = $\frac{8}{27}$

$\Rightarrow$$3^{3n-3m}(9-1)$ = $\frac{8}{27}$

$\Rightarrow$$3^{3n-3m}$ = $\frac{1}{27}$

$\Rightarrow$${27}^{n-m}$ = ${27}^{-1}$

$\Rightarrow$$m-n$ = 1