In this question, the OP asked to find the solutions of: $$a^{−x}+{\log x \over \log a}=0$$ In my answer, I showed that when $a<e^e$ there can be at most one solution. It is also clear that any solution must lie in $x\in[0,1]$.
How can one prove that there exactly three solutions for all $a>e^e$?
Put $\alpha = \log a > e$ and $f(x) = a^{-x} + \frac{\log x}{\log a} = e^{-\alpha x} + \frac{1}{\alpha} \log x$. Then \begin{align*} f(0) &= -\infty \\ f(1/\alpha) &= \frac{1}{e} - \frac{\log \alpha}{\alpha} > 0 \\ f(1/e) &= \frac{1}{e^{\alpha/e}} - \frac{1}{\alpha} < 0\\ f(1) &= e^{-\alpha} > 0 \end{align*} (using the fact that $x < e^x -1$ for positive $x$ in the estimate for $f(1/e)$). Thus $f$ has at least three zeros.
The derivative $$f'(x) = - \alpha e^{-\alpha x} + \frac{1}{\alpha x}$$ vanishes exactly at $x = -\frac{1}{\alpha} W(-\frac{1}{\alpha})$, where $W$ is the Lambert $W$-function. Since $\alpha > e$, there are (despite the notation) two such $x$, which implies that $f$ has at most three zeros.