Finding an exponential equation solution (general trinomial problem) 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 $ax+b-cx^p=0$
 A: Multiplying by $x^p$, you are looking for the zero('s  ?) of the function
$$f(x)=c x^p-a x-b$$ which, for the most general case, will not show explicit solutions. So, either numerical methods or more or less accurate approximations
The first derivatives are
$$f'(x)=c p x^{p-1}-a \qquad \text{and} \qquad f''(x)=c (p-1) p x^{p-2}$$ The fist derivative cancels at
$$x_*=\left(\frac{a}{c p}\right)^{\frac{1}{p-1}}\implies f''(x_*)=c (p-1) p \left(\frac{a}{c p}\right)^{\frac{p-2}{p-1}}$$ So, for a first approximation, assuming that $f(x_*)<0$, expand $f(x)$ around $x_*$ to $O\left((x-x_*)^3\right)$ and solve the quadratic to have
$$x_0=x_*+\sqrt{-2\frac{f(x_*)}{f''(x_*)}}$$
Trying for $a=3$, $b=7$, $c=11$, $p=\pi$,this would give $x_0=1.20223$ while the solution, given using Newton method, is $x=1.03249$.
Now, repeat Newton iterations according to
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 1.20223 \\
 1 & 1.05941 \\
 2 & 1.03333 \\
 3 & 1.03249
\end{array}
\right)$$ which is quite fast because we have easily obtained a reasonable estimate to start with.
You also could improve the guess expanding $f(x)$ around $x_0$ and use series reversion to have
$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}-\frac{f(x_0)^2 f''(x_0)}{2 f'(x_0)^3}+\frac{f(x_0)^3 \left(f^{(3)}(x_0) f'(x_0)-3 f''(x_0)^2\right)}{6 f'(x_0)^5}+\cdots$$ which, for the worked example, would give $x_1=1.03451$.
