Is $x=n^x$ solvable where $n>1$?

Question

Since the graphs of $y=x$ and $y=2^x$ (or $y=n^x$ for that matter, where $n>1$) do not intersect, is $x=2^x$ unsolvable? Or is there some kind of a really clever way to find a root?

P.S. I'm not too sure if this is the right tag to use.

Answer 1

$$x=n^x=e^{x\ln n}$$ $$(-x\ln n)e^{-x\ln n}=-\ln n$$ $$-x\ln n=W(-\ln n)$$ $$x=-\frac{W(-\ln n)}{\ln n}$$ Since $W$ is real-valued on at least one branch when its argument is a small negative number (down to $-\frac1e$), $x=n^x$ has real solutions when $1<n\le e^{1/e}$.