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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.

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    $\begingroup$ See en.wikipedia.org/wiki/Lambert_W_function $\endgroup$
    – user223391
    Commented Jun 13, 2015 at 5:34
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    $\begingroup$ You probably need to be more flexible about what "solve" means. $\endgroup$
    – user14972
    Commented Jun 13, 2015 at 8:23
  • $\begingroup$ This question is answered, though. :) $\endgroup$
    – Sam
    Commented Jun 13, 2015 at 8:25
  • $\begingroup$ What's up with those stars? $\endgroup$
    – Carsten S
    Commented Jun 13, 2015 at 13:48
  • $\begingroup$ @CarstenSchultz they're multiplication symbols $\endgroup$
    – Sam
    Commented Jun 13, 2015 at 21:54

2 Answers 2

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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

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

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  • $\begingroup$ Thanks! Even though I don't fully understand the Lambert function that well (I'll need a formal definition to understand it) I understand how you came up with the solution for a. Agan, Thank! $\endgroup$
    – Sam
    Commented Jun 13, 2015 at 7:20
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    $\begingroup$ The inverse function of $xe^x$ is it's formal defintion. $\endgroup$
    – user2520938
    Commented Jun 13, 2015 at 7:55
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    $\begingroup$ So W(x) is literally just the inverse of $x*e^x$? Ok, I get it all now! $\endgroup$
    – Sam
    Commented Jun 13, 2015 at 10:33
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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.

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