What is a proof that $\ln(\alpha)$ is transcendental for rational number $\alpha$. I believe I heard somewhere that the natural logarithm of any rational number is transcendental. Do you guys have any proofs of that statement?

  • $\begingroup$ Couldn't $\ln(\alpha)$ be irrational but not transcendental? You proved that $\ln(\alpha)$ is irrational, not transcendental. $\endgroup$
    – Aleksandar
    Jul 13, 2015 at 23:10
  • 2
    $\begingroup$ Yes, I deleted my comment for that reason. The transcendence of ln($\alpha$) follows from the Lindemann-Weierstrass theorem en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem $\endgroup$
    – lulu
    Jul 13, 2015 at 23:14
  • $\begingroup$ @lulu I was writing the same as you as an answer. $\endgroup$
    – ajotatxe
    Jul 13, 2015 at 23:16
  • $\begingroup$ Would I be correct in assuming the first sentence should end "for $\alpha$ rational" ? $\endgroup$
    – user88319
    Jul 14, 2015 at 3:03

1 Answer 1


There's a very nice theorem due to Lang reproduced in Appendix 1 of his Algebra from which the Hermite-Lindemann theorem follows as a Corollary. Assuming Hermite-Lindemann which says that if $\alpha$ is algebraic over $\mathbb{Q}$ then $e^{\alpha}$ is transcendental, it follows pretty quickly that $ln(\alpha)$ is transcendental for rational $\alpha$, since $e^{ln(\alpha)}$ is rational. (If $ln(\alpha)$ were algebraic, $e^{ln(\alpha)} = \alpha$ should be transcendental by Hermite-Lindemann.)

  • $\begingroup$ I would be very interested to know if this could be extended to show that $\ln (e+q)$ is transcendental for all $q\in\mathbb{Q},q\neq 0$. If I can retrace my steps I think the three exponentials conjecture may follow. $\endgroup$ Oct 30, 2017 at 8:30
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    $\begingroup$ @RobertFrost The Hermite-Lindemann theorem provides no information regarding $\ln(e+q)$ as $e^{\ln(e+q)}=e+q$ is transcendental, so you can't apply the contrapositive here. $\endgroup$
    – sharding4
    Oct 30, 2017 at 19:16

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