# If $x$ is rational, can $\log(1-x)/\log x$ be algebraic?

If $x$ is positive rational number less than $\frac{1}{2}$, can the following logarithmic expression be equivalent to real algebraic number, say $g$?

$$\frac{\log(1-x)}{\log x} = g$$

• Hint: $\frac{\log(1-x)}{\log x}=g\Rightarrow \log(1-x)=\log x^g\Rightarrow 1-x=x^g$ Jun 4, 2015 at 12:58

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