How to solve $x a^x + b^x = c$?

Question

I need to solve this equation for $x$: $x a^x + b^x = c$ where $a, b, c$ are real numbers. Does it have closed form solution?

Answer 1

In a comment you asked for $a \,2^x - x = c$.

So, consider that you look for the zero(s) of function $$f(x)=a \,2^x - x - c$$ $$f'(x)=a\, 2^x \log (2)-1$$ $$f''(x)=a \,2^x \log ^2(2)$$

The first derivative cancels at $x_*$ $$x_*=-\frac{\log (a \log (2))+1}{\log (2)}$$ which only exist if $a >0$. At this point $$f(x_*)=\frac{\log \left({a \log (2)}\right)}{\log (2)}+\frac{1}{\log (2)}-c$$ $$f''(x_*)=\log (2) > 0$$ If $f(x_*) <0$, then there are two roots which will be given interms of Lambert function $$x_1=-\frac{W_0\left(-a\, 2^{-c} \log (2)\right)}{\log (2)}-c\qquad \text{and} \qquad x_2=-\frac{W_{-1}\left(-a\, 2^{-c} \log (2)\right)}{\log (2)}-c $$