# For which $x$ is $e^x$ rational? Transcendental?

Apart from the trivial cases, $x=\log a$ where $a\in\mathbb{Q}$, are all values of $e^x$ irrational? Are some transcendental?

• they are irrational by definition. Oct 24, 2014 at 13:42
• All transcendental numbers are irrational, so I'm not sure what the role of the "or" is here. Did you mean to ask if $e^x$ is transcendental whenever $x\ne\log{a}$ for some $a\in\mathbb{Q}$?
– mdp
Oct 24, 2014 at 13:52
• I've updated the question. Hope that helps.
– gone
Oct 24, 2014 at 13:57
• @pbs Thanks, that's clearer.
– mdp
Oct 24, 2014 at 14:13
• Would $0$ count? Aug 11, 2015 at 19:58

$e^x$ is rational if and only if $x = \log a$ for $a \in \mathbb{Q}$. This is basically by definition since $\log x$ is defined to be the inverse of $e^x$: $e^x = a$ with $a$ rational $\iff$ $x = \log e^x = \log a$ with $a$ rational.
Similiarly, $e^x$ is transcendental if and only if $x = \log a$ for $a$ transcendental, by the same proof.
An immediate consequence of the Hermite-Lindemann Transcendence Theorem is that if $x$ is algebraic (which includes "rational") and $x\not =0$ then $e^x$ is transcendental.