Original:
I'm trying to solve the following for $x_1$ and $x_2$,
$$ y_1 = x_2 - v\, e^{-x_1} $$ $$ y_2 = x_1 - \frac{1}{v}\, e^{-x_2} $$
in terms of $y_1$, $y_2$, and $v$, which are known and real, and where $v$ is positive. This should be determined and the form seems familiar, but I can't find an inverse. Does anyone know how to go about solving them? Does a solution even exist?
Note 1:
These were simplified from the original relations,
$$ \ln{\gamma_1^∞} = 1 - \ln\left( \frac{V_2}{V_1} \exp{-\frac{\tau_{2,1}}{T}} \right) - \frac{V_1}{V_2} \exp{-\frac{\tau_{1,2}}{T}} $$
$$ \ln{\gamma_2^∞} = 1 - \ln\left( \frac{V_1}{V_2} \exp{-\frac{\tau_{1,2}}{T}} \right) - \frac{V_2}{V_1} \exp{-\frac{\tau_{2,1}}{T}} $$
where $\gamma_1^∞$, $\gamma_2^∞$, $V_1$, $V_2$, and $T$ are positive and $\tau_{1,2}$ and $\tau_{1,2}$ are real. Above, $$v = \frac{V_1}{V_2},\, a=\frac{\tau_{1,2}}{T},\, b=\frac{\tau_{2,1}}{T},\, y_i = \ln{\gamma_i^∞} - (1 + \ln \frac{V_i}{V_{3-i}}) $$
Note 2:
Substituting and reducing yields the equivalent problem:
$$ (x - y)c^{\exp(-x)} = z $$
(where $x = x_1, y=y_2, c=\exp(v), z=e^{-y_1}/v$).
Can this equation be solved for x?
This looks very much as the expressions of the binary activity coefficients at infinite dilution from Wilson or T-K-Wilson models.
I have been facing the same problem as you long time ago and, unfortunately, I did not find any way to solve it (except, as you did, eliminating one of the $x$ and be left with one single equation in terms of the remaining variable). To tell the truth, I was hoping to get solutions in terms of generalized Lambert function. This is probably why were generated nomograms for this class of models.
Then, I gave up and just used numerical methods.
But, you must also know that there could be multiple solutions to the problem.