# Solve $x^2 = 2^x$. [duplicate]

One can see that the solutions are $x=2, 4$ and $x=-0.77$(approximately) seen from the graph. I am posting this to find if there is a way to solve this and find solutions like polynomial equations. Thanks !!!
The trick here is the Lambert W function, an inverse of $we^w$. We have $$x=\sqrt{2}^x=e^{x\ln\sqrt{2}},\,-x\ln\sqrt{2}e^{-x\ln\sqrt{2}}=-\ln\sqrt{2},\,x=-\frac{-W\left(-\ln\sqrt{2}\right)}{\ln\sqrt{2}}.$$ That's the theory, anyway. Like complex exponentiation, the Lambert W-function has multiple branches in $\mathbb{C}$. But the Lambert W function is of interest because many equations' solutions are expressible in terms of it.