I'm looking to solve $a^x + b^x + c^x = 1$ for $x$ when $a, b, c \in [0, 1]$.
If $a+b+c>1$ the solution would be $x>1$, if $a+b+c<1$ then $x < 1$.
A closed formula would be best, but I actually want to do this in JavaScript, so algorithms are also helpful.
Consider that you look for the positive zero of function $$f(x)=a^x + b^x + c^x - 1$$ Its first derivative is always negative and the second derivative always positive.
The series expansion around $x=0$ is $$a^x + b^x + c^x=3+\sum_{n=1}^\infty \frac {\log^n(a)+\log^n(b)+\log^n(c)}{n!} x^n$$ Using a few terms and series reversion, we have $$x=t-\frac{ \log ^2(a)+\log ^2(b)+\log ^2(c)}{2 (\log (a)+\log (b)+\log(c))} t^2+O(t^3)$$ where $t=-\frac {4}{\log(abc)}$.
Let us try with $a=\frac 1 \pi$, $b=\frac 1 e$, $c=\gamma$. This would give, as an estimate, $x=1.25923$ while the "exact" solution given by Newton method is $1.28486$.
We could do better at the price of more terms and very lond expressions. But, starting from this guess, Newton method would work like a charm. $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.259229442 \\ 1 & 1.284569396 \\ 2 & 1.284861903 \\ 3 & 1.284861941 \end{array} \right)$$
