<s>[INST] 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
 [/INST] A hint to get you started:
$$3^{x+2}\cdot 4^{-(x+3)}+3^{x+4}\cdot 4^{-(x+3)}=4^{-(x+3)}\cdot\left(3^{x+2}+3^{x+4}\right)=\frac{1}{4^{x+3}}\cdot\left(3^{x+3-1}+3^{x+3+1} \right)$$
Now try to factor out a term with $3^{(\ldots)}$ and see if you can apply some exponent laws.
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