How to solve the equation $(1-e^{-αa})(1-e^{-βa}) = 1/2$ for a?

Question

Here is what I tried: $(1-e^{-αa})(1-e^{-βa}) = 1/2$ implies $e^{-ln(2)} -e^{-βa} -e^{-αa} + e^{-a(β+α)} = 0$
But from here I don't know how to get the value of "a", any help ?

Answer 1

With Lagrange reversion:

$$(x^r-1)(x-1)=\frac12\implies x=\frac12+\sum_{n=1}^\infty \frac1{n!}\left.\frac{d^{n-1}(x^{r+1}-x^r)^n}{dx^{n-1}}\right|_{x=\frac12}$$

Using the binomial theorem, substituting $x=\frac12$, factorial power $u^{(v)}$, and the regularized Gauss hypergeometric $_2\tilde {\text F}_1$ function gives:

$$(x^r-1)(x-1)=\frac12\implies x=\frac12+\frac12\sum_{n=1}^\infty\sum_{m=0}^n\binom nm(rn+m)^{(n-1)}\frac{(-1)^{n-m}}{n!} 2^{n-rn-m}=\boxed{\frac12+\frac12\sum_{n=1}^\infty\frac{(rn)!(-1)^n}{n!2^{(r-1)n}}\,_2\tilde{\text F}_1\left(-n,rn+1;(r-1)n+2;\frac12\right)}$$

Shown here and both sums are interchangeable when $m$’s upper limit is $\infty$.