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. Then there exist two integers $a,b$ such that $$\left(\frac{a}{b}\right)^{a/b}=2 \\ \frac{a}{b} = 2^{b/a}.$$ But that's impossible because the RHS is rational only for $a=1$, which actually makes it also integer, while with $a=1$ the LHS is non-integer for all $b>1$. Checking that $(a,b)=(1,1)$ yields the false identity $1=2$ concludes the proof.
