Solving the equation $\frac{\ln (x)}{\ln (1-x)} = \frac{1}{x} - 1$

Question

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?

Answer 1

Let $f(x)=x \ln x$. Then the given equation is $$f(x)=f(1-x).$$ This is symmetric about $x=1/2$. Hence $x=1/2$ is a solution.

Answer 2

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.

Answer 3

Let $g:I\to\mathbb{R}$, where $I:=\left[-\dfrac12,+\dfrac12\right]$, be the function defined by $$g(t):=\begin{cases}\left(\dfrac{1}{2}+t\right)\,\ln\left(\dfrac12+t\right)-\left(\dfrac{1}{2}-t\right)\,\ln\left(\dfrac12-t\right)&\text{if }t\in\left(-\dfrac12,+\dfrac12\right)\,,\\0&\text{if }t\in\left\{-\dfrac12,+\dfrac12\right\}\,.\end{cases}$$ (Observe that $g$ is continuous on the whole $I$, and is smooth on $\left(-\dfrac12,+\dfrac12\right)$.) Then, we are to solve for $x\in(0,1)$ from $$g\left(x-\dfrac12\right)=0\,,$$ which is equivalent to solving for $y\in \left(-\dfrac12,+\dfrac12\right)$ such that $$g(y)=0\text{ by setting }y:=x-\dfrac12\,.$$

We claim that $g$ has only three roots $-\dfrac12,0,+\dfrac12$, and this means the only $y\in\left(-\dfrac12,+\dfrac12\right)$ such that $g(y)=0$ is $y=0$, making $x=\dfrac12$ the sole solution to the required equation. Suppose contrary that $g$ has more than three roots. Then, by symmetry, on the interval $\left[0,\dfrac12\right]$, $g$ has at least three roots. Using Rolle's Theorem, $g'$ has at least two roots on $\left(0,\dfrac12\right)$, and so $g''$ has at least one root on $\left(0,\dfrac12\right)$. However, we have $$g'(t)=\ln\left(\frac12+t\right)+\ln\left(\frac12-t\right)+2\,,$$ $$g''(t)=\frac{1}{\frac{1}{2}+t}-\frac{1}{\frac{1}{2}-t}\,,$$ and $$g'''(t)=-\frac{1}{\left(\frac{1}{2}+t\right)^2}-\frac{1}{\left(\frac12-t\right)^2}<0$$ for all $t\in\left(0,\dfrac12\right)$, which is a contradiction we seek.