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Wolfram says $(\ln 2)^2$ is transcendental. I think it says numbers of the form $(\ln a)^b$ are all transcendental, at least for integer $a$ and $b$, I didn't check further.

Maybe there is some corollary from Lindemann's theorem that says something about my question or powers of $\log'$s.

I searched briefly on google for some literature on the irrationality/transcendence on powers of logarithms, either papers or forums, but didn't find anything. Any help would be appreciated.

Pinteco
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    You can forget about the exponent as transcendence is invariant under taking integral roots. The questions becomes if the natural logarithm of an integer greater than or equal to $2$ is trancendental.

    This is answered affirmatively here:

    https://math.stackexchange.com/questions/46497/is-lnn-transcendental-for-all-integer-n-1

     – quid Oct 07 '18 at 00:57

1 Answers1

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Suppose $(\ln a)^b$ is algebraic. There exists a nonzero polynomial $p(x)\in\mathbb Q[x]$ such that $p\left((\ln a)^b\right)=0$. Let $q(x)=p(x^b)\in\mathbb Q[x]$. Then, $q$ is nonzero and $q(\ln a)=0$. So, $\ln a$ is algebraic.

Now if $a$ is a positive algebraic number besides $1$, then it follows from the Lindemann-Weierstrass theorem that $\ln a$ is transcendental. We conclude that $(\ln 2)^2$ is transcendental.

Karl Kroningfeld
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