Is there ANY possible way to solve this equation?

Question

So I came up with this equation and it just seems like I can't solve it AT ALL for '$a$'

$$a*b^a = c$$

EDIT: By the way, I'm only taking $b^a$, not both $b$ and $a$, just in case anyone was confused. Obviously, I would've put parentheses around $a$ and $b$. Just trying to be as specific as possible.

I'm not sure why I can't solve it. I put it into Desmos graphing calculator in the form of '$f(x) = x*a^x$' where '$a$' is some number. There's no point where it's undefined and it is plausible function. So is there ANY way to solve this? Obviously you can't just solving it by using only logarithms.

Answer 1

The Lambert W function is defined as the inverse function of $\color{Blue}{x}e^{\color{Blue}{x}}$. It is a transcendental special function that does not have a closed form in terms of elementary functions. Using it, we have

$$\begin{array}{ll} & a\,b^a=c \\ \iff & a\,e^{(\ln b)a}=c \\ \iff & \color{Blue}{(\ln b)a}\,e^{\color{Blue}{(\ln b)a}}=c\ln b \\ \iff & (\ln b)a=W(c\ln b) \\ \iff & a=\frac{1}{\ln b}W(c\ln b). \end{array} $$

Answer 2

The explicit solution is given by Lambert function, just as whacka answered.

If you cannot use (or don't want to use) this function, then only numerical methods would solve the problem. But first of all, in the real domain, you must know that $W(x)$ is only defined if $x \geq -\frac 1e$ and that it is is double-valued on if $ -\frac 1e\leq x\lt 0 $.

The Wikipedia page dedicated to Lambert function gives formulas for approximating $W(x)$ for given $x$ as well as Newton and Halley iterative schemes for fast and reliable solutions.