How to solve an equation with variable exponents? I have an equation of the form
$$x^a+(1-x)^b = 1$$
where $a$ and $b$ are constants.
Is there some way to manipulate this equation to solve for $x$?
Below is some more information about the problem for some context, although I don't think it affects the mathematics from the above equation.
I am looking at mixing relations for the electrical resistivity in porous media. Usually, we use something called Archie's Law:
$$\rho_b = \rho_f \phi ^{-m}$$
where $\rho_b$ is the bulk resistivity, $\rho_f$ is the fluid resistivity, $\phi$ is the porosity, and $m>0$ is a cementation exponent. In my case, the porosity, $\phi$, is unknown. For the above equation, it is easy to manipulate to solve for $\phi$. However, this simplistic equation is not sufficient for all problems and there is an alternative known as Modified Archie's Law which takes the form:
$$\rho_b\rho_m\phi^m + \rho_b\rho_f(1-\phi)^p = \rho_f\rho_m$$
This form incorporates the surrounding rock resistivity (i.e. the matrix resistivity, $\rho_m$) and uses an additional exponent, $p$. Now, if I want to solve for $\phi$, it is not clear how I can manipulate the equation.
Any help is appreciated.
 A: No, there is no general solution.
Because when $a,b$ are integer the equation is polynomial and it is well known that analytical solutions do not always exist.
A: You may not always be able have other solutions besides those mentioned by lulu in their comments (when $a,b>0$, of course).
In the sequel, I presume, for well-definedness, that $x\in(0,1)$—allowing for $x\in[0,1]$ when $a,b>0$. For our particular demonstration, we consider the general equation $$f(x)=mx^a+n(1-x)^b\,,$$ where $m,n$ are of the same sign—we’d assume positive— and $a,b\notin\{0,1\}$. For now, assume $a=b$. Differentiating $f$ gives $$f’(x)=a(mx^{a-1}-n(1-x)^{a-1})$$ and $$f’’(x)=a(a-1)(mx^{a-2}+n(1-x)^{a-2})\,.$$ This shows that $x=\frac{c}{1+c}$, where $c:=\left(\frac{n}{m}\right)^{1/(a-1)}$ is a global minimum (respectively, maximum) of $f$ if $|a-\frac{1}{2}|> \frac{1}{2} $ (respectively, $a\in(0,1)$), which in either case is simply $$f(\frac{c}{1+c})=\frac{mc^a+n}{(1+c)^a}\,.$$ For $a<0$, in which case it is a minimum, it follows that $f(x)\ge 2$ when $m=n=1$ and there’s no solution to your equation (more generally, there’s no solution when the minimum is $>1$). On the other hand, it is easy to observe that before and beyond $x= \frac{c}{1+c}$, then $f$ is strictly monotonic, which implies that besides $x\in\{0,1\}$ —when $m=n=1$—we cannot have that $f(x)=1$ (and this is generally the case if the maximum is $>1$).
From the above, owing to continuity — and as probably hinted in G Cabs comment— it follows that a small perturbation of $b$ from $a$ would not alter the solution set as obtained above; in particular, there’d exist $\epsilon$ (may or may not depend on $a,b$) such that whenever $|a-b|\le\epsilon$, then $f(x)=1$ has no solutions when the case with $a=b$ also has no solutions.
