An improved model for the deadtime of a radiation detector is given by $$m = \frac{n e^{-n \tau_P}}{1 + n \tau_N} $$ where n is the assumed true count rate, m the measured count rate, and $\tau_P$ and $\tau_N$ are paralyzable and non-paralyzable deadtimes. But if you're making a measurement and want to make a correction to estimate the true count rate, that equation has to be inverted. I thought the Lambert W function might help. As I understand it, given $y=x e^x$, then $W(y) = x$, but we need the same x on the top and bottom. So, $$-m\tau_P = \frac{-n\tau_P}{1+n\tau_N}e^{-n\tau_P}$$ That seemd like a good start, but what about the denominator? Try raising both sides to the power of $\dfrac{1}{1+n\tau_N}$, $$(-m\tau_P)^{1/(1+n\tau_N)} = \left(\frac{-n\tau_P}{1+n\tau_N}\right)^{1/(1+n\tau_N)} e^{\frac{-n\tau_P}{1+n\tau_N}}$$ So close, but that power has to be applied to everything or nothing. Is there a way forward from here?
Edit, based on the answer by Claude Leibovici:
I had trouble working from equation (4) since I didn't find an explicit solution. But in section 2, the second and third equations below equation (5), seemed to be the heart of it. In my case, with one term on the top and one on the bottom, $$ e^x\frac{x-t}{x-s}=a$$ is solved by $$x=W\left({t\atop s};a\right)$$ Putting my equation in something like that form gives $$e^{-n\tau_P}\frac{n}{n\tau_N-(-1)}$$ Define $x=-n\tau_P$, so $n=-x/\tau_P$, do some factoring, multiply both sides by a $\tau_N$, and I get $$e^x\frac{x}{x-\frac{\tau_P}{\tau_N}}=m\tau_N$$ Then, identifying $t=0, s=\frac{\tau_P}{\tau_N}, a=m\tau_P$, the solution is $$n=-\frac{1}{\tau_P}W\left({0\atop \tau_P/\tau_N};m\tau_N\right)$$ The solution is Theorem 1, which states (and using $i$ because I already have an $n$) $$W\left({t \atop s};a\right)=\frac{\tau_P}{\tau_N}\sum_{i=1}^{\infty} \frac{L'_i\left(i\frac{\tau_p}{\tau_N} \right)}{i}(m\tau_N)^i$$ where $T=t-s$ and $L'_i$ is the derivative of the $i$th Laguerre polynomial. Since $t=0$, $e^{-i t}=1$. And that gets rid of that stupid $\tau_P$ in the denominator, which is a small correction and should cause no disasters if $\tau_p=0$. $$n=-\frac{1}{\tau_N} \sum^{\infty}_{i=1} \frac{L'_i\left(i\frac{\tau_P}{\tau_N}\right)}{i}(m\tau_N)^i$$
But the minus sign... counts per second shouldn't be negative...
2nd edit: I think I figured out the minus sign. The first few terms of the sum are $$-m\tau_N - m^2(\tau_N^2-\tau_N\tau_P)+m^3(stuff), \tau_P<\tau_N$$
