Some special cases of the Lambert W function

Suppose $f(x)$ is given by the functional relation $$\begin{equation} e^{-a(x) f(x) + b(x)} = c(x) f(x) + d(x) \end{equation}$$ then $$\begin{equation} f(x) = \frac{1}{a(x)}W\left( \frac{a(x)}{c(x)} e^{b(x)+\frac{a(x)d(x)}{c(x)} } \right) - \frac{d(x)}{c(x)} \end{equation}$$ where $W$ is the Lambert W Function. You can click through to Wikipedia for a proof. Here I merely list some special cases which every good citizen should instantly recognize (and it can be mildly tedious to reproduce them).

Recovering the definition of $W(x)$

Just to check we are on the right planet set $a=1$, $c=1$, $b=\ln(x)$ and $d=0$ to get $$\begin{equation} x e^{-f(x)} = f(x) \end{equation}$$ or equivalently $f(x)e^{f(x)} = x$ which is the definition of the Lambert W function. Sure enough those substitutions recover $$f(x) = W(x)$$ as expected. A minor modification uses $a(x)=1$, $c(x)=1$, $b(x)=0$ but leaves $d(x)$ general.

The solution to $e^{-f(x)} - f(x) = d(x)$

is $$\begin{equation} f(x) = W \left( e^{d(x)} \right) \end{equation}$$ Only slightly more generally, set $a(x)=a$, $b=\ln(g(x)), c=1, d=1$ for those cases where $g(x)$ is on the wrong side, as it were.

The solution to $e^{-af(x)} = f(x) g(x)$

or equivalently $f(x) e^{af(x)} g(x) = 1$ is $$\begin{equation} f(x) = \frac{1}{a} W\left( a g(x) \right) \end{equation}$$ In particular,

The solution to $x^k e^{-af(x)} f(x) = 1$

which seems to crop up a fair bit for your author is $$\begin{equation} f(x) = \frac{1}{a} W\left( a x^{-k} \right) \end{equation}$$

Similarly setting $a=-1$ ...

The solution to $x e^{f(x)} f(x) = 1$

(i.e. where $x$ is on the wrong side but we otherwise have the Lambert W definition) must be $$\begin{equation} f(x) = - W\left( -\frac{1}{x} \right) \end{equation}$$ We might also take $b = \ln(g(x))$ and thus

The solution to $g(x) e^{-af(x)} = f(x)$

or equivalently $f(x)e^{af(x)} = g(x)$ is $$\begin{equation} f(x) = \frac{1}{a} W\left( a g(x) \right) \end{equation}$$ which reduces to the Lambert W function for $a=1$, as we expect. It is also pretty obvious from first principles, because if we multiply both sides by $a$ we have $$\begin{equation} (af) e^{(af)} = ag \end{equation}$$ and thus $af = W(ag)$. Next suppose we want a power of $f$ to appear. Let $b=\beta(x)/k$, $c = \gamma^{1/k}$, $a = k/\alpha$ and $d=0$. And then raise both sides to the power $k$. It follows that...

The solution to $e^{\alpha f(x) + \beta(x)} = \gamma f(x)^k$

is $$\begin{equation} f(x) = \frac{\alpha}{k}W\left( \frac{k e^{\frac{1}{k}\beta(x)} }{\alpha \gamma^{1/k}} \right) \end{equation}$$ and if, in particular, $\beta(x) = -\ln(g(x))$ then

The solution to $e^{\alpha f(x)} = g(x) \gamma f(x)^k$

is $$\begin{equation} f(x) = \frac{\alpha}{k}W\left( \frac{k }{\alpha g(x)^{1/k} \gamma^{1/k}} \right) \end{equation}$$ and if we take $c = \frac{1}{\gamma}$ and $g(x)=x$ and $\alpha = \frac{s}{2}$ and $k=2$ then p>

The solution to $e^{-\frac{s}{2}f(x)} x f(x)^2 = c$

is $$\begin{equation} f(x) = \frac{s}{2k}W\left( \frac{2k}{s} \sqrt{ \frac{c}{x}} \right) \end{equation}$$