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

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

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

$3 \cdot 9^n = 2 \cdot 4^n $

$ 3^{2n+1} = 2^{2n+1}$

$(\frac{3}{2})^{2n+1} = 1$

So, $2n+1 = 0 $

$ n = -\frac{1}{2}$

Start by dividing top and bottom by $4^n$: $$\frac{\frac{9^{n+1}}{4^n} + \frac{4^{n+1}}{4^n}}{\frac{9^n}{4^n} + 1} = 6.$$ Substitute $u = \frac{9^n}{4^n}$ to get $$\frac{9u + 4}{u + 1} = 6 \iff 9u + 4 = 6(u + 1) \iff u = \frac{2}{3}.$$ Therefore, $$\left(\frac{9}{4}\right)^n = \frac{2}{3} \implies n = -\frac{1}{2}.$$

Hint $\,\ \overbrace{\{ 9^{\large n+1},\, 4^{\large n+1}\} = \{6\cdot9^{\large n}\!,\, 6\cdot4^{\large n}\}}^{\large \text{by equal sum & product}}\,\ $ so $\,\ 9^{\large n+1}\! = 6\cdot 4^{\large n}\Rightarrow\, \left(\dfrac{3}2\right)^{\!\large 2n}\!\! = \dfrac{2}3$