Is there an explicit solution to $a^x+b^x=1$? Where $a, b \in [0, 1]$ and $a+b \le 1$.
I've been playing around with this equation, but I can't seem to make any progress in solving it. I tried it in Wolfram for some hints, but nothing showed up. Should I assume there isn't an explicit solution to this?
There are certain pairs of points ${(a,b)}$ for which we can find a nice analytical solution. As an example, take ${a=\frac{1}{2}, b=\frac{1}{4}}$ (so ${b=a^2}$). Then let ${u=\left(\frac{1}{2}\right)^x}$. The equation ${\left(\frac{1}{2}\right)^x + \left(\frac{1}{4}\right)^x = 1}$ simplifies to
$${u + u^2 = 1\Leftrightarrow u^2 + u - 1 = 0}$$
Now you can find the roots of this to find the value of ${\frac{1}{2^x}}$, and hence rearrange to find the exact value of ${x}$.
In general, if ${b = a^{\frac{p}{q}}}$, then let ${u=a^x}$:
$${\Rightarrow u + u^{\frac{p}{q}} = 1}$$
Then let ${u = t^q}$
$${\Rightarrow t^q + t^p = 1}$$
So in this case where one of the numbers ${a,b}$ can be written as a rational power of one another - you can reduce the equation to a sort of "polynomial" - this again, provides no promises of a closed form solution, but depending on the choice of ${p,q}$ it might.
To summarize: in general, I don't think you are going to find a nice closed form. However, if a "nice" relationship between ${a}$ and ${b}$ exists - it's possible a closed form can be found by substitutions and reductions.
For a graphical solution, change variables to $x=\log_a(u)$ to get $u+b^{\log_a(u)}=1$ and treat the second term on the left-hand side as an independent variable (for now) by setting $\log_b(v)=\log_a(u)$ to get $$u+v=1 $$ which you have to simultaneously solve under the condition $$u=a^{\log_b(v)}$$ which stems from $u,v=u(x),v(x)$.
What this does is that it gives you a very simple plot for the first equation and the "hard" part is in plotting the second one. But once you've done this, the intersection of the two plots will give you the approximate answer you're looking for.
I wrote it in slightly different way: $$a^x+b^x=1 \Leftrightarrow a^{\log_{a}\alpha}+b^{\log_{b}(1-\alpha)}=1$$ so $x = \log_{a}\alpha = \log_{b}(1-\alpha)$ and here is possibility to construct solution for given $0<a,b, a+b=1$ taking $0<\alpha<1$, because there is always intersection on $O\alpha x$ plane.
May be it will be helpful/interesting property of these solutions $\log_{a}b=\log_{\alpha}(1-\alpha).$
WLOG, assume that $0 < b < a < 1$. Denote $p = \frac{\ln a}{\ln b} \in (0, 1)$. With the substitution $u = b^x$, the equation becomes $u^p + u = 1$ which admits an infinite series solution (see [1]) $$u = \sum_{k=0}^\infty \frac{\Gamma(pk+1)(-1)^k}{\Gamma((p-1)k+2) k!}.$$
Also, we may use the Lagrange inversion theorem to obtain infinite series solutions (see [2], [3]).
Reference
[1] Nikos Bagis, Solution of Polynomial Equations with Nested Radicals, https://arxiv.org/pdf/1406.1948.pdf
[2] https://mathoverflow.net/questions/118532/what-was-lamberts-solution-to-xmx-q
[3] https://en.wikipedia.org/wiki/Lagrange_inversion_theorem
