Working out the number of solutions for $x\in\Bbb R$ in $$a^x+b^x=c$$ with $a,b,c\in\Bbb R^+$ and $a\neq b$, I found the following: The number of solutions is $\def\artanh{\operatorname{artanh}}$
$$S(\alpha,\beta) = \begin{cases} 1, & \text{if } |\alpha|>1 \\ 1, & \text{if } |\alpha|=1 \text{ and } \beta>-\ln2\\ 0, & \text{if } |\alpha|=1 \text{ and } \beta\leqslant -\ln2 \\ 0, & \text{if } |\alpha|<1 \text{ and } \beta < g(\alpha) \\ 1_2,&\text{if } |\alpha|<1 \text{ and } \beta = g(\alpha) \\ 2, & \text{if } |\alpha|<1 \text{ and } \beta > g(\alpha) \\ \end{cases}\tag 1$$ with $$ \alpha = \frac{\ln(a\cdot b)}{\ln(a/b)}, \qquad \beta = \ln(c/2) \tag 2 $$ and $$ g(x) = -\frac12 \ln \left(1 - x^2\right) - x \artanh x \tag 3$$
$1_2$ stands for "one solution of multiplicity 2", and solutions without subscript are of multiplicity 1.
From what I can tell (1) is correct, but it appears to me that it's more complicated than needed, so I'd like to simplify it; in particular the expressions involving $g(\alpha)$. So far I have the following simplifications:
The relation $\beta\sim-\ln2$ can be simplified to $c\sim 1$ with $\sim\,\in\{>,\leqslant\}$.
The relation $|\alpha|\sim1$ is $\alpha^2 \sim 1$ which becomes $\ln a\ln b\sim 0$.
But the most complicated is of course $\beta \sim g(\alpha)$, where I get a mess after substituting $\alpha$.