If $x$ is positive rational number less than $\frac{1}{2}$, can the following logarithmic expression be equivalent to any real number , say $r$?
$$\frac{\log(1-x)}{\log x} = r$$, Where $r$ is positive real number less than one,
Assume first $r$ is a rational number say $n/m$, then you would get this equation, $x^n=(1-x)^m$, where solution for $x$ will be clearly algebraic - irrational number contradicting the assumption of rationality of $x$, hence $r$ cannot be rational number & must be irrational number, if so, you get this following equation, $x^r = 1-x$, the right hand side is a rational number, but the left hand side is transcendental number according to - Gel-schneider theorem, that contradicts again the assumption of irrationality of $r$, so what is that number suppose to be?
I guess someone will simply say it is a transcential number given by definition, if so assume $x$ an integer power of a rational number, say $x=a^p$, where $p$ is odd prime number, then by substitution you get this simple equation which is a reduced form of FLT for rational numbers , $a^p + (a^r)^p = 1$, but we already know that $a^r = b$, where $b$ is a rational number that doesn't exist from the proof of Andrew Wiles & Taylor to Fermat's Last Theorem, thus $log(b) / log(a)$ is not Transcendental number, it is a kind of numbers that have to be redefined again & precisely