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?
 A: If $x$ is rational, $p/q$, then $2^{q/p}$ is rational. That's only possible if $p=1$ and $p/q$ is an integer.
A quick way to write it, assuming $p,q$ are relatively prime:
$$\left(\frac pq\right)^{p/q}=2\implies p^p=2^qq^p$$
But $p$ and $q$ are relatively prime, using unique factorization, we see that $p^p,q^p$ are relatively prime, so $q^p=1$ which means either that $p=0$ (not possible) or $q=1$. If $q=1$, then $p^p=2$, so again by unique factorization, $p=2^k$ and $1=k2^k$, which can't be solved.
A: 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.
A: 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?
