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 $A>0$ (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 $y<1$ the left hand side is negative, whereas the right hand side is positive and conversely, for $y>1$ the left hand side is positive, whereas the right hand side is negative. Hence $y=1$ is the only solution to your equation.