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?

  • 1
    $\begingroup$ Sidenote: It can relatively easily be shown that this is equivalent to solving $x^x=\left(1-x\right)^{\left(1-x\right)}$ but I'm not sure if that leads anywhere. $\endgroup$
    – Jam
    Commented Dec 7, 2018 at 13:58

3 Answers 3


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.

  • 4
    $\begingroup$ Very nice solution! $\endgroup$
    – user
    Commented Dec 7, 2018 at 8:17
  • 5
    $\begingroup$ I think this is insufficient. Proving that $x=\dfrac12$ is a solution is easy. Proving that it is the only solution is not as easy, but I can see that one may approach it by showing that $$f(x)-f(1-x)<0\text{ for all }x\text{ such that }0<x<\frac12\,,$$ and by symmetry $$f(x)-f(1-x)>0\text{ for all }x\text{ such that }\frac12<x<1\,.$$ $\endgroup$ Commented Dec 7, 2018 at 8:41
  • 2
    $\begingroup$ Yep, IMO too many upvotes as the existence of other solutions should be discussed. $\endgroup$
    – user65203
    Commented Dec 7, 2018 at 12:53

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.

enter image description here


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.


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