Question

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

Answer 1

$\frac{9^n\times 3^2\times {(3}^{-\frac{n}{2}}{)}^{-2}-{27}^n}{3^{3m}\times 2^3}=\frac{1}{27}$
$\frac{9^n\times 9\times 3^n-{27}^n}{3^{3m}\times 8}=\frac{1}{27}$
$\frac{9^m\times 3^m\times 9-{27}^n}{3^{3m}\times 8}=\frac{1}{27}$
$\frac{{(9\times 3)}^n\times 9-{27}^n}{8\times {27}^m}=\frac{1}{27}$
$\frac{{27}^n\times 9-{27}^n}{8\times {27}^m}=\frac{1}{27}$
$\frac{(9-1){27}^n}{8\times {27}^m}=\frac{1}{27}$
$\frac{8\times {27}^n}{8\times {27}^m}={27}^{-1}$
${27}^{n-m}={27}^{-1}$
Therefore by comparing powers,we get
$n-m=-1$
$m-n=1$
Hence proved