$\log \log (\alpha)$ transcendental??

Question

$\log \log (\alpha)$ transcendental?? ($\alpha$ algebraic $\neq 0$ and $1$)

I supposed $\log \log (\alpha)=\beta$ , $\beta$ transcendental. Then $\log(\alpha)=e^{\beta}$ and it is know $e^{\beta}$ is transcendental.

Answer 1

I don't know if we can prove it right now, but it is likely.

An interesting conjecture by Schanuel states that

If $x_1,\dotsc,x_n$ are complex numbers linearly independent over $\Bbb{Q}$, then $$ \text{trdeg}_{\Bbb{Q}}(x_1,\dotsc,x_n,e^{x_1},\dotsc,e^{x_n}) \geq n $$

If this conjecture holds then $$ 2 \leq \text{trdeg}_{\Bbb{Q}}(\alpha, \log(\alpha), \log(\alpha), \log\log(\alpha)) = \text{trdeg}_{\Bbb{Q}}(\log(\alpha), \log\log(\alpha)) \leq 2 $$ so $\log(\alpha)$ and $\log\log(\alpha)$ would be algebraically independent for every $\alpha \in \bar{\Bbb{Q}} \setminus \{0,1\}$.

Answer 2

The answer is NOT. We are sure $\log\alpha$ is trascendental because $\alpha$ is algebraic neither $0$ nor $1$ but the log of a trascendental may be either algebraic or trascendental ($\log e=1$ and $e$ is trascendental)