Prove that $e^{\frac{1}{\log(x)}}$ is at least countably transcendental

Question

Q: Prove that $f(x)=e^{\frac{1}{\log(x)}}$ is at least countably transcendental for $x\in\Bbb R\cap (0,1).$

Answer 1

Let $f(x)=e^{\frac{1}{\log(x)}}$ for $x\in\Bbb Q~\cap(0,1).$ $f(x)=x\implies$ that $1=\ln^2(x),$ which is never true for $x\in\Bbb Q~\cap(0,1).$ This implies that $x$ cannot be rational in the first place and therefore must be irrational. Letting $x$ be an algebraic irrational number $\in (0,1)$ and doing $f(x)=x\implies1=\ln^2(x),$ which is never true for irrational algebraic $x\in(0,1).$ This implies that $x$ cannot be irrational algebraic in the first place, and therefore must be transcendental, for at least one value $x\in(0,1).$ This transcendental value is $x=1/e.$

Now let $f(x)=x^n$ for $n\in\Bbb N.$ By the same logic $x$ must be transcendental for all $n\in \Bbb N.$ This implies that $f(x)$ contains at least a countable number of transcendental values. The same logic shows that $x$ must be transcendental for $n$ rational and $n$ irrational algebraic.