Question

$\frac{9^{n+1}\times {\left(3\frac{-n}{2}\right)}^{-2}-{27}^n}{{(3^m\times 2)}^3}=\frac{1}{729}$
Prove that $m-n=2$

Answer 1

First let us simplify the left side of the given equation,

$\begin{array}{cc}LHS& =\frac{9^{n+1}\times {\left(3^{-\frac{n}{2}}\right)}^{-2}-{27}^n}{{\left(3^m\times 2\right)}^3}\\ & =\frac{{\left(3^2\right)}^{n+1}\times 3^{-\frac{n}{2}\times -2}-{\left(3^3\right)}^n}{{\left(3^m\right)}^3\times {\left(2\right)}^3}\\ & =\frac{{3^2}^{n+2}\times 3^n-3^{3n}}{3^{3m}\times 2^3}\\ & =\frac{{3^3}^{n+2}-3^{3n}}{3^{3m}\times 2^3}\\ & =\frac{3^{3n}\left(3^2-1\right)}{3^{3m}\times 8}\\ & =3^{3\left(n-m\right)}\end{array}$

Now we will rewrite the right side of the given equation,

$\begin{array}{cc}RHS& =\frac{1}{729}\\ & =\frac{1}{3^6}\\ & =3^{-6}\end{array}$

On comparing $LHS$ and $RHS$ we get,

$\begin{array}{cc}3\left(n-m\right)& =-6\\ n-m& =-2\\ m-n& =2\end{array}$

Hence, proved.