Show that for $\;\dfrac12I have the problem of unique. The existence is follows since using sided LHospital
$$\begin{cases}\lim\limits_{x\to0^+}f(x)=1\\{}\\\lim\limits_{x\to1^-}f(x)=\dfrac12\end{cases}\implies \;\text{the IVT for continuous functions proves}$$ .
For unique I did: evaluate differential
$$f'(x)=-\frac1{x\log^2x}+\frac1{(x-1)^2}=\frac{x\log^2x-(x-1)^2}{x(x-1)^2\log^2x}$$
Now, if I show $\;f'<0\;$ then $\;f\;$ is monotone going down and there is unique solution, but I'm having lots of problem for this: the denominator is always positive, but I can't prove numerator is negative.
After lots of tryings I tried direct proof:
$$f(x)=f(y)\iff \frac{\log\dfrac yx}{\log x\log y}=\frac{x-y}{(x-1)(y-1)}$$
and I won't have way to show $\;x=y\;$ .
Any helps is greatly appreciated.
 A: In order to prove that
$$ f(x)=\frac{1}{\log x}-\frac{1}{x-1}$$
is decreasing over $(0,1)$, we just need to prove that:
$$ g(t) = \frac{1}{t}-\frac{1}{e^t-1} $$
is decreasing over $\mathbb{R}^-$. It is sufficient to prove that:
$$ \forall t<0,\quad g'(t) = \frac{e^t}{(1-e^t)^2}-\frac{1}{t^2}<0,$$
or:
$$ \forall t<0,\quad e^t+e^{-t}>2+t^2 $$
that is equivalent to:
$$ \forall t<0,\quad \cosh t > 1+\frac{t^2}{2} $$
that is trivial, since for any $t$:
$$ \cosh t = \sum_{n\geq 0}\frac{t^{2n}}{(2n)!} = 1+\frac{t^2}{2}+\frac{t^4}{24}+\ldots.$$
A: We need $x\log^2 x-(x-1)^2<0$ for $0<x< 1$.
It's like $\frac{1}{x}\log^2 \frac{1}{x}-(\frac{1}{x}-1)^2<0$ for $x>1$.
$$\begin{aligned}\frac{1}{x}\log^2 \frac{1}{x}-(\frac{1}{x}-1)^2<0
\Leftrightarrow\\ \log^2 \frac{1}{x}-x(\frac{1}{x}-1)^2<0
\Leftrightarrow\\ (-\log x)^2<\frac{(1-x)^2}{x}
\Leftarrow \log x<\frac{x-1}{\sqrt{x}}\end{aligned}$$
$t=\sqrt{x},x=t^2, \log x=2\log t, x>1\Rightarrow t>1$
$$2\log t<\frac{t^2-1}{t}=t-\frac{1}{t}\text{ for }t>1$$
LHS=RHS=0 when $t=1$ and $\frac{2}{t}=(2\log t)'<\left(t-\frac{1}{t}\right)'=1+\frac{1}{t^2}$ when $t>1$.
A: Let us consider
$$f(e^x)=f(e^y)\implies\dfrac{1}{x}-\dfrac{1}{e^x-1}=\dfrac{1}{y}-\dfrac{1}{e^y-1}$$ with $x,y\in(-\infty,0).$
$$y(e^y-1)(e^x-x-1)=x(e^x-1)(e^y-y-1)$$
After simplifying this expression you can obtain $$(y-x)(1-e^x)(1-e^y)+xy(e^y-e^x)=0$$ Suppose $y\gt x\gt0.$  Then $(y-x)(1-e^x)(1-e^y)+xy(e^y-e^x)\gt0.$
Therefore this expression is zore if and only if $$\color{Green}{x=y.}$$
A: $x.(\log x)^2$ has limit 0 near 0 because we can write it as  ($\log x)^2/(1/x)$ and use L'Hospital's Rule to replace it by $-2 \log x/(1/x)$ and use the same rule again to get the 0 limit.
