I recently read this problem Solve $9^x+6^x=2^{(x+1)}$ which had the obvious solution of $x=0$ as well as a simple proof that no other solution was possible.
But how would you solve examples such as $9^x+6^x=2^{(x+2)}$, $6^x+4^x=3^{(x+2)}$, $9^{(x-1)}+2^{(x+5)}=4^{(x+3)}$ (two solutions) or even $6^x+4^x=3^{(x+1)}+2^x?$
The simple answer is that there is no nice way of solving these equations involving exponentials, except for some carefully crafted equations with obvious solutions, i.e, $9^x+6^x=2^{x+1}$.
However, for some equations, you may be able to tell the number of real solutions. For example, for the equations, $9^x+6^x=2^{x+2}$ and $6^x+4^x=3^{x+2}$, write them as $$\left(\frac{9}{2}\right)^x+3^x=4$$ and $$2^x+\left(\frac{4}{3}\right)^x=9.$$ For both equations, the left hand side is an increasing function while the right hand side is a constant function, so there can only be one real solution.
Similarly, you can do some analysis on similar equations by writing them in a similar form as the two equations I have shown but you can't find explicit expressions for most of these equations.