Solve $3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$ Can someone point me in the right direction how to solve this?
$3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$
I guess I have to get to logarithms of the same base. But how? What principle should I use here?
Thx

Using $a^{-b}=\frac{1}{a^b}$, The equation can be rewritten as 
$$\frac{3^{x+2}}{4^{x+3}}+\frac{3^{x+4}}{4^{x+3}}=\frac{40}{9}$$
$$\implies \frac{9}{64}\left(\frac{3}{4}\right)^x+\frac{81}{64}\left(\frac{3}{4}\right)^x=\frac{40}{9}$$
$$\implies \left(\frac{3}{4}\right)^x=\frac{256}{81}$$
$$\implies x=-4$$
