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

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

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.

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No chance unless there is some arcane special function you can write it in terms of. – Brendan McKay Apr 4 '12 at 7:39
@ Brendan - Are you teasing the OP?! Looks like homework to me. Differentiate twice and you've got the solution no? – Kevin Smith Apr 4 '12 at 10:11
Generally, the solution to the derivative of an equation is not a solution to the equation itself. – Gerald Edgar Apr 4 '12 at 12:52

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? ]

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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:

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

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Why is this question community wiki? – Kevin Smith Apr 4 '12 at 16:11
Questions automatically become CW if they have more than a certain number of revisions. Details are available in the site documentation somewhere. I don't know whether that feature accounts for what happened here. – Gerry Myerson Apr 5 '12 at 4:45