If x is rational, can log⁡(1−x)log⁡x be algebraic?

Your identity gives:
${\displaystyle 1-x=x^g}$
where ${\displaystyle x\in \mathbb{Q}}$ trivially gives that the LHS is a rational number. If ${\displaystyle g}$ is not a rational number, the Gelfond-Schneider theorem gives that the RHS is a trascendental number, contradiction.
So ${\displaystyle g}$ has to be a rational number. But in order that ${\displaystyle 1-x}$ and ${\displaystyle x^g}$ are rational numbers with the same denominator, ${\displaystyle g}$ has to be one. So ${\displaystyle x=\frac{1}{2}}$ and ${\displaystyle g=1}$ is the only solution.