Lambert W function equation I was attempting to solve different types of equations using the Lambert W function. I stumbled across one that depending on how the signs were placed I was able to solve it or not. For instance, I’ve had no problem solving $e^{-2x} + 1 = 3x$, however, solving $e^{-2x} - 1 = -3x$ has been somewhat imposible. I’ve tried all types of substitutions and transformations albeit nothing seems to work. If someone could guide me on how to solve the equation that will really help me out.
Thanks!

Multiplying by $e^{2x}$ both sides gives
$$1-e^{2x}=-3xe^{2x}$$
$$e^{2x}(3x-1)=-1$$
$$e^{2x}(2x-\frac23)=-\frac23$$
$$e^{2x-\frac23}(2x-\frac23)=-\frac23 e^{-\frac23}$$
$$2x-\frac23=W_n(-\frac23 e^{-\frac23})\,\,\, n\in\mathbb{Z}$$
$$x=\frac13+\frac12W_n(-\frac23 e^{-\frac23})\,\,\, n\in\mathbb{Z}$$
There are infinite values of the Lambert-W function and in this case there are two real valued solutions - the principal value of the Lambert-W function $W_0(-\frac23 e^{-\frac23})=-\frac23$ and the value $W_{-1}(-\frac23 e^{-\frac23})\approx-1.429355228 $
