I have an equation of the form $$ax^{1-p}+bx^{-p}=c$$where $x,a,b,c,p\in \mathbb{R}_+$ and $p\ge 1$. $a,b,c,p$ are given constants. I'm looking for clues for solving it. Does it even permit any closed form solution?
I was able to deduce the following. We introduce $f(x) = ax^{1-p}+bx^{-p}-c$
$$\lim_{x\to 0^+}f(x) = \frac{ax+b-cx^p}{x^p}\rightarrow+\infty$$
and $\lim_{x\to +\infty}f(x)=-c\le 0$. Furthermore $f'(x) = a(1-p)x^{-p}-bpx^{-p-1} = x^{-p-1}[a(1-p)x-bp]\le 0$, therefore according to intermediate value theorem there is a unique root somewhere in $\mathbb{R}_+$ since the function is continuous except in $x=0$. BTW I tried it with Wolfram-alpha but it was unable to understand (which indicates that maybe I don't know how to interact with it)! How should we proceed?
Thanks in advance
===========================Edit===========================
As suggested by Gary, it is possible to get other forms of this equation by changing variable. for example if
We introduce $z=\frac{1}{x}$, we get $az^{p-1}+bz^p=c$.
We can write $x^p = e^{p\ln x}$ and define $t=\ln x$ and get $ae^{(p-1)t}+be^{pt}=c$
Also multiplying both sides with $x^p$ gives $ax+b-cx^p=0$