Solve $-B \ln y -A y \ln y + A y- A =0$ for $y$

I would like to know if there is a (preferably closed-form) solution for

$-B \ln y -A y \ln y + A y- A =0$ for $y$

Where $A, B \in \mathbb{R}^{+}$. I have reasons to think there isn't a closed form solution, but I am also sure that there are solutions other than $y=1$

Is there any theorem saying there are no closed-form solutions to this equation? If there is, do you think an approximate solution might be possible? If there isn't, any ideas how to solve it?

(I already asked this question before, but with a sign change that made the solution almost trivial. It was a typo, so I'm asking again)

Note: My question is motivated by the need for a solution to this problem:

$1-\frac{A}{x} \gamma (2,\frac{x}{B}) = 0$ for $x$

The solution needn't be exact, but it'd be great if it was in closed form (I'm trying to avoid numerical approximations). Especially, I'm interested in the case $r$ around $B$, but also would like convergence for $B \rightarrow 0$.

My approach was to perform an asymptotic expansion on the incomplete lower gamma function following http://dlmf.nist.gov/8.11, truncating the series after 2 terms. After simplifying, Mathematica gives me:

$x + x \frac{A}{B} e^{-\frac{x}{B}}+A e^{-\frac{x}{B}} -A =0$

So, after the substitution $x=-B \ln y$, I get: $-B \ln y -A y \ln y + A y- A =0$ for $y$

Maybe there is a better approach to this problem?

Thanks!

Solution 1:

Hint:

$$0=-B\log y-Ay\log y+Ay-A=\log\frac{1}{y^B}+\log\frac{1}{y^{Ay}}+Ay+A=\log\frac{1}{y^{Ay+B}}+Ay+A$$

This doesn't look, as you say, like having a closed form but perhaps the above, together or not with xavier's comment (Lambert function), can give you some good approximation.