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If $a\in\mathbb{R}\setminus\left\{0,1\right\}$ is an algebraic number, can $\ln\left(a\right)$ ever be a Liouville number?

This is not a homework question, nor do I know much about the innards of proving these kinds of things. I am just very interested in transcendental numbers.

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    $\ln(a)$ is known to be transcendental for $a$ algebraic, and of course Liouville numbers are known to be transcendental. Combined with the continuity of $\ln(x)$ over $\Bbb R^+$ and that the Liouville numbers form a dense, uncountable subset of $\Bbb R$, I feel like the answer is "probably." Granted this is mostly handwaving and is far from anything resembling a proof. – PrincessEev Apr 03 '19 at 06:05

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Almost certainly it can't, but I would be surprised if this were provable in the current state of the art, even in the case where $a$ is rational.

Robert Israel
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The answer to your question is no. This is the case because the exponential of each Liouville number is transcendental. See my paper here.It is open access.

Bumblebee
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SID MORRIS
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    -1: not only is this a link-only answer, the paper in the link, restricted to the case of $\log_e(\text{algebraic})$, itself is a "link-only paper", referring to statements in Mahler's papers. – ronno Apr 05 '23 at 12:07
  • This answers the question completely using Mahler sets. What more could you ask? – SID MORRIS Apr 08 '23 at 09:36
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    What is the point of linking your paper when you know that all your paper says about this question is "this is a theorem by Mahler from this other paper"? At the very least link/refer to Mahler's paper(s) directly. As presumably an expert on the topic, you could also add a summary of the argument. It would make sense if your paper was included as an additional reference for related results. – ronno Apr 08 '23 at 09:45