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It is known how to solve equations of the form $e^{-cx}=a_0(x-r)$ in terms of the Lambert $W$-function. Namely, according to Wikipedia, the solution is given by $$ x=r+\frac{1}{c}W\bigg(\frac{ce^{-cr}}{a_0}\bigg). $$

Is there a similar way to solve equations of the form $$ e^{-cx^2}=a_0(x-r), $$ where say $a_0,c,r>0$? In this case, there is indeed a unique solution with positive $x$ value, for the left hand side is positive and the right hand side is linear and increasing in $x$ and is negative at $x=0$. Without the $r$, one could reduce to the above solution by squaring. I'd be interested in if there are either special functions that are reasonably well-understood to solve this, or if there are any reasonable approximations on the value of $x$ in terms of $c,a_0,$ and $r$. Thanks!

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  • $\begingroup$ Well, the $r=0$ case reduces to the Lambert function if we square both sides. The general case is similar to generalized Lambert functions that invert $R(x)e^x$ for a rational $R$, except here it is only rational in $\sqrt{x}$. Did this come up in an application? $\endgroup$ – Conifold Jun 27 at 5:07
  • $\begingroup$ No, this is not solvable. $\endgroup$ – Simply Beautiful Art Aug 9 at 13:56

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