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The question is in the title "Is there an algebraic $a$ such that $\ln(a)$ is algebraic other than 1?" It was inspired by this question.

I don't see an easy answer, but might be missing something.

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DRF
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    My suggested duplicate target asks about natural numbers in place of $a$, but the answers cover the algebraic case as well. Please search the site before asking/answering. – Jyrki Lahtonen Feb 27 '17 at 12:55
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    @JyrkiLahtonen Does it really make sense to mark as duplicate a question which asks something completely different (which btw is why I didn't find it) even if the answer turns out to work? This is a bit like saying that the question "Why are there no zero divisors in R" and "Why does there exist a unique inverse for every real number" are the same since "Because R is a field" answers both. – DRF Feb 27 '17 at 13:21
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    Point taken. But IMHO whoever wants to ask or answer this question should first search the site a bit more thoroughly. I don't have the time to look for an exact duplicate, but that is really not my job. – Jyrki Lahtonen Feb 27 '17 at 14:03
  • But since you asked nicely, see this, this, this. To be fair, the users involved in the question that inspired you to ask this one should have done their part searching the site also. – Jyrki Lahtonen Feb 27 '17 at 14:37
  • @JyrkiLahtonen I bow to your searching prowess (and I don't mean that sarcastically). None of those actually ask about algebraic numbers but all of them have answers pointing to Lindemann-Weirstrass (or similar) which is admittedly the essential point. All in all Yeap I should have searched better. – DRF Feb 27 '17 at 14:45
  • I confess that I used Lindemann as a search word, so it was more difficult for you as you had not heard of the result. The answerers OTOH could have done the same if so inclined. – Jyrki Lahtonen Feb 27 '17 at 14:47

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This is the same as asking for an algebraic number $b\ne 0$ such that $e^b$ is algebraic.

Such a number cannot exist, due to the Lindemann-Weierstrass theorem: Because $b\ne 0$, the set $\{b\}$ is linearly independent over $\mathbb Q$, so the theorem says that $\{e^b\}$ is an algebraically independent set over $\mathbb Q$; in particular $e^b$ is transcendental.

hmakholm left over Monica
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