TITLE: Confusion regarding usage of Lambert function
QUESTION [6 upvotes]: I stumbled upon an equation that goes like:
$$e^{\pi x} - \frac{x}{k} = -1$$
I learnt that Lambert function is useful when dealing with such equations where it can take the form $f(x) = xe^x$.
So, the equation essentially becomes:
$$ x = \frac{1}{\pi} \ln\Big(\frac{x}{k} - 1\Big)$$
Is there any way that I can make  $\ln\Big(\frac{x}{k} - 1\Big)$ as some $e^{f(x)}$ so that I can use the Lambert function?
Any other way of solving the equation is always welcomed :) Thanks.

REPLY [5 votes]: We begin with the expression
$$e^{\pi x}-x/k=-1$$
Upon rearranging, we find that
$$\begin{align}
\pi k&=e^{-\pi x}(\pi x-\pi k)\\\\
&=e^{-\pi(x-k+k)}(\pi x-\pi k)\\\\
-\pi ke^{\pi k}&=e^{-\pi (x-k)}(-\pi(x-k))
\end{align}$$
Can you finish now?

REPLY [2 votes]: You can rearrange as
$$
ke^{\pi x}=x-k\\
k=(x-k)e^{-\pi x}\\
-\pi k=-\pi(x-k)e^{-\pi x}\\
-\pi k=-\pi(x-k)e^{-\pi(x-k)}e^{-\pi k}\\
-\pi ke^{\pi k}=-\pi(x-k)e^{-\pi(x-k)}
$$
now set $w=-\pi(x-k)$ and $z=-\pi ke^{\pi k}$, so you have
$$
z=we^w.
$$