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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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    My suggested [duplicate target](http://math.stackexchange.com/q/15285/11619) asks about natural numbers in place of $a$, but the answers cover the algebraic case as well. Please search the site before asking/answering. – 2017-02-27
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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. – 2017-02-27
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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. – 2017-02-27
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    But since you asked nicely, see [this](http://math.stackexchange.com/a/1360098/11619), [this](http://math.stackexchange.com/q/989096/11619), [this](http://math.stackexchange.com/q/46497/11619). 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. – 2017-02-27
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    @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. – 2017-02-27
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    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. – 2017-02-27

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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.