How to solve for x when $x^x=2x$

Question

How do you solve for $x$ when $x^x=2x$? Graphing both $y^x=y+x$ and $y=x$, there are two intersection points, one at $x=2$, and another at $x\approx0.34$. I've recently learned the Lambert-W function, but haven't been able to apply it to this problem.

Answer 1

As said in comments, there is no solutions in terms of Lambert function and, then, as already suggested, Newton method is probably the simplest to be used.

Looking for the zero of $$f(x)=x^x-2x$$ $$f'(x)=x^x (\log (x)+1)-2$$ and using $x_0=\frac 12$, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.5 \\ 1 & 0.33573216085622873 \\ 2 & 0.34626771116191797 \\ 3 & 0.34632336075977158 \\ 4 & 0.34632336227858092 \end{array} \right)$$