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?

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    $\begingroup$ The only access we have to most real numbers is by approximating them using rational numbers. Thus the only access we have for the solutions for most equations is approximating them using rational numbers. Approximation is always a method, and usually the only method. $\endgroup$ – OR. Sep 12 '13 at 3:09

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.

  • $\begingroup$ Good enough for me, thank you! Just learned what a Lambert W function is $\endgroup$ – Yadnarav3 Sep 12 '13 at 3:34

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