# Is the logarithm of base transcendental irrational for every integer greater or equal than two?

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.

• Well, yes, since $\log_b n=\frac pq$ means $b^p-n^q=0$. – user228113 Feb 3 '17 at 22:43
• @G.Sassatelli Would they be also transcendental? – Nick Cassol Feb 3 '17 at 22:45

$\log_b n = \frac{p}{q}$ 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.