closed form solution to $x\ln (\frac{2}{x})=k(a-x) \ln (\frac{2}{a-x})$

Question

$x\ln (\frac{2}{x})=k(a-x) \ln (\frac{2}{a-x})$

where $a$ and $k$ are positive constants. $a$ is usually small, say, $0< a<0.1$ and $x\in (0,a)$.

There are ways to calculate numerical solutions. However, as it is a intermediate step of an optimization problem, I do need closed form results to move forward with some strict proof. Does that exist?

One observation is that function $f(x)=x\ln (\frac{2}{x})$ is increasing when $x\in (0,2/e)$, thus with $0< a<0.1$ and $x\in (0,a)$, $f(x)$ is increasing. I don't know if this helps.

Answer 1

Rather than solving for $x$ in terms of $a$, how about solving for $a$ in terms of $x$ like this: $$ a = x + \frac{x \operatorname{ln}(x/2)}{k \mathrm{W}\left(\frac{x \operatorname{ln} (x/2)}{2 k}\right)} $$ where $\mathrm{W}$ is the Lambert W function.
[Inspired by the solution to What is the name of $\frac{e^z-1}{z}$ and how to invert it? ]

Answer 2

This isn't a research level mathematics question, but you do have to do a bit of high school calculus. Since $y\log(2/y)$ is strictly increasing and $k(a-y)\log(2/(a-y))$ is strictly decreasing on $(0,a)$, dividing the l.h.s by the r.h.s, differentiating twice and setting the result to $0$ should give you the unique point where the curves meet.

EDIT: This is incorrect (or at least it only holds for $k=1$, which trivially gives you the mid-point) because the inflection point is unchanged by the constant $k$. Apologies.

Speaking of arcane special functions! the $q$-digamma function function might be of some use:

$$\frac{d}{dz} \log\Gamma_q(z)=-\log(1-q)+\log(q)\sum_{0}^{\infty}\frac{q^{n+z}}{1-q^{n+z}}.$$