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 or an approximation that is valid when $r$ is in the order of $B$ and $B \rightarrow 0$.

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 hint how to solve it?



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!
Since $A&gt;0$ (the only solution for $A=0$ would be $B=0$ and $y$ arbitrary anyway) we can divide by $A$ to obtain
$$(y+\frac{B}{A})\ln y = 1-y$$
Note that $y=1$ is certainly a solution to your equation. Furthermore, note that for $y&lt;1$ the left hand side is negative, whereas the right hand side is positive and conversely, for $y&gt;1$ the left hand side is positive, whereas the right hand side is negative.
Hence $y=1$ is the only solution to your equation.
I doubt that the equation has a nontrivial solution. Implicitly differentiating once,

$$
B/y + 2A + A\ln y = 0.
$$

Rearranging,

$$
\mathrm{e}^2y = \mathrm{e}^{-B/Ay}.
$$

Implicitly differentiating once more,

$$
\mathrm{e}^2 = -\frac{B}{A}\mathrm{e}^{-B/Ay}
$$

which seems to suggest that $y$ is constant.