I want to solve for $x$ the equation
$$ x^{\alpha} (1-x)^{1-\alpha} = (\gamma x)^{\beta} (1-\gamma x)^{1-\beta} $$
where $\alpha,\beta,\gamma, x$ are all strictly between zero and one.
If I'm not mistaken, this is a transcendental equation and an analytic solution may not exist. I however would like to get as close as possible to a closed form solution of the form $x = RHS$.
What assumptions do I need to simplify the problem? Would making $\alpha,\beta,\gamma$ rational help?
In case a solution can't be found explicitly, could I still find some way to describe how the $x$ for which the quality holds increases or decreases as a function of the parameters?
This answer uses $a,b,c$ for MathJax and readability. Rearrange: $$x^a(1-x)^{1-a}=(cx)^b(1-cx)^{1-b}\iff f(x)=(1-cx)^{b-1}(1-x)^{1-a}x^{a-b}=c^b$$
For $a<b$, $g(x)=f^{-1}(x)$is monotonically decreasing and contains $(0,1),(\infty,0)$:
We apply Mellin inversion. $g(x)$’s Mellin transform is:
$$\text M_s(g(t))=\int_0^\infty g(t)t^{s-1}dt=\int_0^1t f(t)^{s-1}df(t)$$
Now apply tabular integration:
\begin{matrix}&\text D&\text I\\+&t&f(t)^{s-1}f’(t)\\-&1&\frac{f(t)^s}s&\\+&0&\frac1s\int_0^1f(t)^sdt\end{matrix}
so the regularized Gauss hypergeometric function integral representation appears:
$$\text M_s(g(t))=-\frac1s\int_0^1 (1-ct)^{(b-1)s}(1-t)^{(1-a)s}t^{(a-b)s}dt=-\frac{\operatorname B((1-a)s+1,(a-b)s+1)}s\,_2\text F_1((1-b)s,(a-b)s+1;(1-b)s+2;c)$$
Finally, apply the inverse Mellin transform. The solution to the transcendental equation is $g(c^b)$.
$$\bbox[3px,border: 3px #00BFFF ridge]{x^a(1-x)^{1-a}=(cx)^b(1-cx)^{1-b}\implies x=\frac1{2\pi i}\int_{d-i\infty}^{d+i\infty}\frac{c^{-sb}}s \operatorname B((1-a)s+1,(a-b)s+1)\,_2 \text F_1((1-b)s,(a-b)s+1;(1-b)s+2;c)ds,a<b}$$
The Mathematica code:
{FindRoot[x^a(1-x)^(1-a)-(cx)^b(1-cx)^(1-b),{x,1}],1/(2\[Pi] I)NIntegrate[c^(-s b)/s Beta[1+s-a s,1+a s-b s] Hypergeometric2F1[s-b s,1+a s-b s,2+s-b s,c],{s,d-I*p,d+I*p}]}verifies the formula for given $a,b,c,p,d$:
Expanding the beta function and hypergeometric function via its Mellin-Barnes integral, makes it resemble a “two parameter Fox H function” which likely gives a “closed form”, but it is not widely accepted as little research on it has been done.