Suppose $a$ and $b$ are reals such that $a^b=b^a$. If $a$ is algebraic, is $b$ algebraic too?
The answer is no. For instance, let $a=3$ and $b\neq 3$ be the real number satisfying $3^b=b^3$. Clearly $b$ is not an integer. It follows that $b$ is irrational -- indeed, if $b$ was a non-integer rational, $3^b$ would be irrational, while $b^3$ would be rational. Finally, $b$ is transcendental, since otherwise $b$ would be algebraic irrational, $b^3$ would be algebraic and $3^b$ would be transcendental by Gelfond-Schneider.