How to solve strange exponential equation?

Question

How would equations of the form

$b^x-x^a=0$ be solved for $x$, given $a$ and $b$? For instance, specifically, how would

$2^x=x^2$ be solved? Does a method exist?

Answer 1

Beginning with $$ b^x=x^a $$ and writing $b^x=e^{x\ln b}$: $$ 1=xe^{-x\frac{\ln b}{a}}. $$ Multiplying both sides by $-(\ln b)/a$ yields $$ -\frac{\ln b}{a}=\left(-x\frac{\ln b}{a}\right)e^{-x\frac{\ln b}{a}}. $$ Therefore $$ W\left(-\frac{\ln b}{a}\right)=-x\frac{\ln b}{a} $$ where $W$ is the Lambert W function and so $$ x=-\frac{a}{\ln b} W\left(-\frac{\ln b}{a}\right). $$ This can be confirmed by Wolfram Alpha.