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?
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Couldn't $\ln(\alpha)$ be irrational but not transcendental? You proved that $\ln(\alpha)$ is irrational, not transcendental. – Aleksandar Jul 13 '15 at 23:10
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2Yes, I deleted my comment for that reason. The transcendence of ln($\alpha$) follows from the Lindemann-Weierstrass theorem https://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem – lulu Jul 13 '15 at 23:14
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@lulu I was writing the same as you as an answer. – ajotatxe Jul 13 '15 at 23:16
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Would I be correct in assuming the first sentence should end "for $\alpha$ rational" ? – Jul 14 '15 at 03:03
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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.)

sharding4
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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. – it's a hire car baby Oct 30 '17 at 08:30
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1@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. – sharding4 Oct 30 '17 at 19:16
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That last comment doesn't seem to follow.
Assume BWOC $\ln(e+q)=a$ is algebraic. Then $e+q=e^a$. But then $e^a-e-q=0$ which is a violation of Lindemann Weierstrauss for any algebraic $a$ and $q\neq 0$.
– Mason Jul 07 '22 at 21:37