TITLE: How to solve $a^x+b^x=1$ (solve for $x$)
QUESTION [19 upvotes]: I am sorry if this is too easy question for this site, but I really can't find the solution...
$a^x+b^x=1$

Tyma Gaidash asked for context:
I don't have much to add, I was trying to understand how fast population grow assuming that everyone born exactly same amount of time (for example everyone have 2 children 1 born 40 years after the parent born and the other 20 years after the parent born, is it faster than having a twins after 28 years)
If I assume the population formula is $a^x$ and I need to find a
I get that the time it will take to the population multiply itself is the solution to the equation:
$\sum_{i=1}^{n}2^{-t_i/x}=1$ where $t_i$ it after how many year children I born after his parent.
I hoped that there is a way to solve this equation (at least for 2 children ) but I haven't found one.

REPLY [9 votes]: In general, there is no analytical solution.
If $$b=a^{m/n}$$ where $m,n$ are naturals, the equation can be written as
$$z^n+z^m=1$$ where $z:=a^{x/m}$. There are explicit formulas for the cases $m,n\le4$, and specific numerical methods for polynomials of higher degree. But in the latter case and when the exponent is irrational, direct resolution with an equation solver (Newton) is simpler.

Another way to look at the problem is by setting $t:=a^x$ so that
$$t^{\log b/\log a}+t=1$$ or $$t^\alpha+t=1.$$
From this
$$\alpha=\frac{\log(1-t)}{\log(t)}=:f(t)$$ gives you an explicit relation between $t$ and $\alpha$.
What you are after is the inverse, $t=f^{-1}(\alpha)$ and you can some how precompute/tabulate this inverse, by interpolation or other means.
We have the pretty good approximation
$$f^{-1}(\alpha)=\frac{(\alpha-1)(\sqrt{0.25\alpha})^\alpha+1}{\alpha(\sqrt{0.25\alpha})^{\alpha-1}+1}$$ for $\alpha<1$.