$$ \huge 4^x + 6^x = 9^x $$

Key is first to identify that for $4,6$ and $9$, we only have two factors, $2$ and $3$. Hence if we divide both sides by $4^x$, we get

$$ \huge 1 + \left(\frac{3}{2} \right)^x = \left(\frac{3}{2} \right)^{2x} $$

We end up with a quadratic equation, solving for $\left(\frac{3}{2} \right)^x$ yields

$$ \huge \left(\frac{3}{2} \right)^x = \frac{1+\sqrt{5}}{2} $$

It follows that

$$ \huge x = \frac{\ln \left( \frac{1+\sqrt{5}}{2} \right)  }{\ln \left( \frac32 \right)} $$