# Is $(\ln 2)^2$ transcendental?

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.

• 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: math.stackexchange.com/questions/46497/… – quid Oct 7 '18 at 0:57

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.