Question

Find $n$
$\frac{9^{n+1}+4^{n+1}}{9^n+4^n}=6$

Answer 1

We have,

$\begin{array}{c}\frac{9^{n+1}+4^{n+1}}{9^n+4^n}\\ 9^n\cdot 9+4^n\cdot 4=6\left(9^n+4^n\right)\\ 9\cdot 9^n+4^n\cdot 4-6\times 9^n-6\times 4^n=0\\ 9^n\left(9-6\right)+4^n\left(4-6\right)=0\\ 3\cdot 9^n-2\cdot 4^n=0\\ 3\cdot 9^n=2\cdot 4^n\\ \frac{9^n}{4^n}=\frac{2}{3}\\ {\left(\frac{3}{2}\right)}^{2n}={\left(\frac{3}{2}\right)}^{-1}\\ 2n=-1\\ n=\frac{-1}{2}\end{array}$

Hence, this is the answer.