TITLE: $2^x$ is irrational if $x$ is irrational?
QUESTION [6 upvotes]: Prove/Disprove that if $x$ is irrational, then $2^x$ is also irrational.
My attempt for the proof:
Suppose $2^x>0$ is a rational number, then $2^x=\frac{a}{b}$ for some natural numbers $a$ and $b$. Taking logarithm with base $2$ on both sides to get, $x=\log_2 \frac{a}{b}$. Here I stuck! how to reach at $x$ is rational?

REPLY [23 votes]: This is false in general, take $x=\frac{\ln 3}{\ln 2}$, then $2^x=e^{x\ln 2}=e^{\ln 3}=3\in\mathbb{Q}$ but $x\notin\mathbb{Q}$. Otherwise, there would exist $p,q\geqslant 1$ coprimes such that $x=\frac{p}{q}$, that is to say $q\ln 3=p\ln 2$ and thus $3^q=2^p$ which is not because $2$ and $3$ are coprimes.