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

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

  • 4
    $\begingroup$ Hint: $\frac{\log(1-x)}{\log x}=g\Rightarrow \log(1-x)=\log x^g\Rightarrow 1-x=x^g$ $\endgroup$ – Scippy Jun 4 '15 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.

  • $\begingroup$ Actually the question was excluding (x=1/2 or x= 0), therefore, no solution for the alleged question or equation unless (g) is transcendental number? $\endgroup$ – bassam karzeddin Jun 4 '15 at 13:38
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    $\begingroup$ @bassamkarzeddin: yes, exactly. $\endgroup$ – Jack D'Aurizio Jun 4 '15 at 13:39
  • $\begingroup$ Which proves FLT in short!! $\endgroup$ – bassam karzeddin Jun 4 '15 at 13:43
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    $\begingroup$ I don't believe this proves FLT... $\endgroup$ – Zach Effman Jun 4 '15 at 22:16

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