Is the solution of the equation $x^x=2$ rational? Let $x$ be the solution of the equation $x^x=2$. Is $x$ irrational? How to prove this?

Suppose $x$ is rational. Clearly $x<2$ and $x>1$, so $x$ is not an integer. Now from this question we obtain that $x^x=2$ is irrational, a contradiction: If $x$ is a positive rational but not an integer, is $x^x$ irrational?
