We have $$4^x + 6^x = 9^x$$
I simplified a bit and got the answer to be something like :
We have
$$\begin{align}{(2^2)}^x + 2^x 3^x &= (3^2)^x\\ \frac{{(2^2)}^x}{{(3^2)}^x} + \frac{2^x}{3^x} &= 1\end{align}$$ So,
$$\frac{2^x}{3^x} = \frac{-1\pm\sqrt{5}}{2}$$
Thus
$$x = \frac{\log\left(\frac{-1\pm\sqrt{5}}{2}\right)}{ \log\frac{2}{3}}$$
Simplifying, $$x = \frac{{\log(\sqrt{5}-1)}- {\log2}}{{\log2}-{\log3}}$$ Am I right or something is wrong?