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.
Is the logarithm of base transcendental irrational for every integer greater or equal than two?
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logarithms
irrational-numbers
transcendental-numbers
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3Well, yes, since $\log_b n=\frac pq$ means $b^p-n^q=0$. – 2017-02-03
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0@G.Sassatelli Would they be also transcendental? – 2017-02-03
1 Answers
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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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0Thank you for your answer, especially the transcendental part! :D – 2017-02-04