I gotta solve
\begin{align}
(\ln(ab) - ab + c)\left(\frac{1}{a} - b\right) = 0
\end{align}
for $a$, where $a,b,c \in \mathbb{R}_+$.
We clearly have a solution at $a = \frac{1}{b}$. However, Matlab tells me that we also have a = -wrightOmega(- c - pi*1i)/b. No idea what I should make of it.
- Can I neglect the second solution due to the fact, that $a$ is supposed to be real?
The first solution, is as you correctly note:
$$a=\frac{1}{b}$$
The second equation can be solved in terms of the Lambert $W$ function, as:
$$ \begin{align} \ln(ab)-ab+c&=0\Rightarrow\\ \ln(ab)&=ab-c\Rightarrow\text{ (pass through $\exp$)}\\ ab&=e^{ab}\cdot e^{-c}\Rightarrow\\ ab\cdot e^{-ab}&=e^{-c}\Rightarrow\\ -ab\cdot e^{-ab}&=-e^{-c}\Rightarrow\\ -ab&=W_k(-e^{-c})\Rightarrow\\ a&=\frac{-W_k(-e^{-c})}{b}\text{, $k\in\mathbb{Z}$}\\ \end{align} $$
Now, the Wright-Omega function $\omega$ satisfies:
$$W_k(z)=\omega(\ln(z)+2k\pi i)\text{, $k\in\mathbb{Z}$}$$
Therefore you can express the solutions in terms of this function as:
$$a=\frac{-\omega(\ln(-e^{-c})+2k\pi i)}{b}\text{, $k\in\mathbb{Z}$}$$
Choosing the principal branch of $\ln$ and for the particular case where $c\ge 1$ the branch of $W$ which gives real solutions, is the $k=-1$ branch, so:
$$ \begin{align} a&=\frac{-W_{-1}(-e^{-c})}{b}\Rightarrow\\ a&=\frac{-\omega(\ln(-e^{-c})+2(-1)\pi i)}{b}\Rightarrow\\ a&=\frac{-\omega(\pi i -c - 2\pi i)}{b}\Rightarrow\\ a&=\frac{-\omega(-c -\pi i)}{b} \end{align} $$