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

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  • $\begingroup$ they are irrational by definition. $\endgroup$
    – Dunham
    Oct 24, 2014 at 13:42
  • $\begingroup$ 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 $a\in\mathbb{Q}$? $\endgroup$
    – mdp
    Oct 24, 2014 at 13:52
  • $\begingroup$ I've updated the question. Hope that helps. $\endgroup$
    – gone
    Oct 24, 2014 at 13:57
  • $\begingroup$ @pbs Thanks, that's clearer. $\endgroup$
    – mdp
    Oct 24, 2014 at 14:13
  • $\begingroup$ Would $0$ count? $\endgroup$
    – Amad27
    Aug 11, 2015 at 19:58

2 Answers 2

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$e^x$ 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.

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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.

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