Solution of $\frac{c - 1}{x - c} + \log \frac{c - x}{x - 1} = 0$ for $x$ I was looking for the maximum of the function  $f(x) = \left(x - 1 \right) \log\frac{c - x}{x - 1}$ for $\{x,c\} \in\mathbb{R}^+$, $x\not=1$ (obviously) and $x \le c-1$. The normal way to find such maxima is to find the zero(s) of the first derivative, here $f'(x) = \frac{c - 1}{x - c} + \log \frac{c - x}{x - 1}$. So far, so good(?).
What I tried myself:
Part the logarithm from the rest and and exponentiate (with $e = \exp(1)$) $$ \frac{c-x}{x-1} = e^{-\frac{c-1}{x-c}}$$ That's messy, so substitute $$u =  -\frac{c-1}{x-c}$$ and $$v = \frac{\left(c - x\right)^2}{\left(c - 1\right) \left(x - 1\right)} $$ to get $$ \frac{u}{e^u} = \frac{1}{v} $$ That gives $$u = -\mathrm{W}\left(-\frac{1}{v}\right)$$ ($\mathrm{W}$ is Lambert's W-function)  with the attached condition $v \not= 0$ which is further restricted to $e \le v $ to get a real solution. But $v$ does not hold the stricter condition  $e \le v $ which gives me the faint idea that I must have made something wrong somewhere, somehow, probably right in the beginning.
I don't even know if an analytic solution exists.

Let $y=\frac{c-x}{x-1}$.  Then, $x=\frac{y+c}{y+1}$ and $\frac{c-1}{x-c}=-\left(\frac{y+1}{y}\right)$.  Therefore, we can write $\frac{c-1}{x-c}+\log\left(\frac{c-x}{x-1}\right)=0$ in terms of $y$ as 
$$\log y=1+\frac1y \tag 1$$
From $(1)$, it is easy to show that
$$\frac1e=\frac1y e^{1/y} \tag 2$$
Inasmuch as the definition of the Lambert W Function is defined by the equation $z=W(z)e^{W(z)}$, solution to $(2)$ is given by
$$\begin{align}
y&=W^{-1}\left(\frac1e\right) \tag 3\\\\
&\approx 3.59111829181573 
\end{align}$$
where the notation $W^{-1}(z)$ denoted the reciprocal of $W(z)$.  Solving $(3)$ for $x$ yields
$$\bbox[5px,border:2px solid #C0A000]{x=\frac{1+cW\left(\frac1e\right)}{1+W\left(\frac1e\right)}}$$
