Confusion regarding usage of Lambert function

Question

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.

Answer 1

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?

Answer 2

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 $$ z=we^w. $$