I am deducing from the Schanuel's conjecture the following statement:
Let $\alpha, \beta$ be positive real algebraic numbers with $\alpha,\beta \neq 1$ and $\frac{\log \beta}{\log \alpha} \notin \mathbb{Q}$. Then $\{x \in \mathbb{R}: \alpha^x, \beta^x \in \overline{\mathbb{Q}}\}=\mathbb{Q}$.
The inclusion $\supseteq$ is easy. For $\subseteq$, after applying the Schanuel's conjecture two times by finding an algebraic relation between the logarithms, I am now at the point that $\log \alpha, \log \beta, \log \alpha^x$ are linearly dependent over $\mathbb{Q}$. Then I write a linear relation
$$A\log \alpha+B\log \beta+C\log \alpha^x=0$$
and then do something to get a contradiction. For this I solved the case $B=0$: in this case we have $A,C \in \mathbb{Q}$ and thus $x$ must be in $\mathbb{Q}$. So we only need to solve the case $B \neq 0$. But then I don't know what to do next: I tried many approaches but led to the dead-end.
Any help is appreciated.
Best Answer
I successfully solved this stuck so I will leave my proof here in case somebody will need it.
So, $\log \alpha^x$ could be represented by a $\mathbb{Q}$-linear combination
$$\log \alpha^x=A_1\log \alpha+B_1 \log \beta.$$
For the argument in the question post, we can reuse it replacing $\log \alpha^x$ by $\log \beta^x$ and get another $\mathbb{Q}$-linear combination
$$\log \beta^x=A_2\log \alpha+B_2 \log \beta.$$
Let $t=\dfrac{\log \alpha}{\log \beta} \notin \mathbb{Q}$, then by dividing both sides of the two linear combinations by $\log \alpha$ and $\log \beta$ respectively, we get $$x=A_1+B_1t=A_2.\dfrac{1}{t}+B_2.$$ Thus $B_1t^2+(A_1-B_2)t-A_2=0$. This quadratic equation, with unknown $t$, has coefficients living in $\mathbb{Q}$, thus $t \in \overline{\mathbb{Q}}$ and then $x \in \overline{\mathbb{Q}}$. Therefore the case $x \notin \mathbb{Q}$ cannot happen since it will make $\alpha^x$ and $\beta^x$ become not algebraic anymore, a contradiction. We conclude that $x \in \mathbb{Q}$.