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I was studying natural logarithms and I stumbled upon the fact that every natural logarithm of a natural number greater than two is irrational, is it the same for every transcendental base? Thanks.

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    $\begingroup$ Well, yes, since $\log_b n=\frac pq$ means $b^p-n^q=0$. $\endgroup$ – user228113 Feb 3 '17 at 22:43
  • $\begingroup$ @G.Sassatelli Would they be also transcendental? $\endgroup$ – Nick Cassol Feb 3 '17 at 22:45
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$\log_b n = \frac{p}{q}$ if $b^p=n^q$. This means that $b^p - n^q = 0$, which, since $p$, $n$, and $q$ are all positive integers here, means that $b$ is the root of a polynomial with integer coefficients, that is, that $b$ is algebraic.

As for $\log_b n$ always being transcendental for transcendental $b$ and integer $n\ge2$, that is false: consider the base $2^\sqrt{2}$. $\log_{2^\sqrt{2}}(2) = \frac{1}{\sqrt{2}}$, and in general if $b=n^k$ for algebraic $k$, $\log_b n = \frac{1}{k}$, which is also algebraic.

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  • $\begingroup$ Thank you for your answer, especially the transcendental part! :D $\endgroup$ – Nick Cassol Feb 4 '17 at 0:29

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