Question

If $\frac{9^n\times 3^2\times 3^n-(27{)}^n}{(3^3{)}^5\times 2^3}=\frac{1}{27}$, find the value of n.

Answer 1

$\frac{9^n\times 3^2\times 3^n-(27{)}^n}{(3^3{)}^5\times 2^3}=\frac{1}{27}$

$\Rightarrow \frac{(27{)}^n\times 3^2-(27{)}^n}{(3^3{)}^5\times 2^3}=\frac{1}{3^3}$

$\Rightarrow \frac{(27{)}^n\times [9-1]}{\left({27}^5\right)\times 2^3}=\frac{1}{27}$

$\Rightarrow \left({27}^{n-5}\right)\times \frac{8}{8}={27}^{-1}$

$\Rightarrow (n-5)=-1$

$\Rightarrow n=5-1=4$

$\therefore n=4$