TITLE: Solving the equation $\frac{\ln (x)}{\ln (1-x)} = \frac{1}{x} - 1$
QUESTION [6 upvotes]: I'm trying to solve the following equation (which has solution $x = 1/2$)  
$$\frac{\ln (x)}{\ln (1-x)} = \frac{1}{x} - 1 $$
I can't seem to do it analytically. Any ideas?

REPLY [6 votes]: The function $x\log x-(1-x)\log(1-x)$  is only defined in $(0,1)$, is differentiable, has a root at $x=\frac12$ (by inspection) and tends to zero at the interval endpoints.
The derivative,
$$\log x+\log(1-x)+2$$ has two roots in $(0,1)$, where $x(1-x)=e^{-2}$.
Hence the function is negative in $(0,\frac12)$, with a single minimum, and positive in $(\frac12,1)$, with a single maximum, and there are no other roots.