Give an equation below: $$ \frac{x^k-a^k}{x-a}=c \qquad (1) $$ where $1<a<x$, $0<k<1$, and $c>0$.
I can easily find the numerical root of (1) by using Newton's method or the other tools.
However, I still want to ask this question:
Can we find the closed-form root of (1) w.r.t. $x$?
Any helpful suggestions or answers will be greatly appreciated!
I've been working in a new special function called Lambert-Tsallis, $W_r(z)$. Such a function is the solution of the following problem
$$W_r(z)\bigg( 1 + \frac{W_r(z)}{r} \bigg)^r= y\cdot e_r(y) = z, \space (1)$$ being $y= W_r(z)$ and $e_r(y) = \bigg( 1 + \frac{y}{r} \bigg)^r$.
There are different math problems that can be reduced to Eq. (1). Examples 1, 2 and so on.
After a few manipulations your original problem can be written as $$ x^r\cdot e_{r}\bigg(-\frac{r}{c}x^{k-1} \bigg) = z^r$$ with $\color{red}{r=k-1}$ and $\color{red}{z=a-\frac{a^k}{c}}$ which produces $$\frac{W_r(-\frac{r}{c}z^r)}{-\frac{r}{c}}=x^r$$ and the final solution of your problem:
$$\color{red}{x=\frac{z}{1+\frac{W_r(-\frac{r}{c}z^r)}{r}}}$$
As Tyma Gaidash writes in the comments, the equation can be transformed to a trinomial-like equation:
$$\frac{x^k-a^k}{x-a}=c$$ $$-cx+x^k-a^k+ac=0$$ $$-\frac{c}{a^k-ac}x+\frac{1}{a^k-ac}x^k-1=0$$ $x\to -\frac{a^k-ac}{c}t$: $$t-\frac{(a^k-ac)^{k-1}}{c^k}t^k-1=0$$ $k\to\alpha$: $$t-\frac{(a^\alpha-ac)^{\alpha-1}}{c^\alpha}t^\alpha-1=0$$ $\frac{(a^\alpha-ac)^{\alpha-1}}{c^\alpha}\to y$: $$t-yt^\alpha-1=0$$
Now the equation is in the form of equation 8.1 of [Belkic 2019]. Solutions in terms of Bell polynomials, Pochhammer symbols or confluent Fox-Wright Function $\ _1\Psi_1$ can be obtained therefore.
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Hint:
Let us define the irrational function $\;f(x)=x^k\;$ with your data, then:
$$\lim_{x\to a}\frac{x^k-a^k}{x-a}=f'(a)=\ldots$$
Or also Lagrange's MVT:
$$\frac{x^k-a^k}{x-a}=f'(d)\;,\;\;\text{for some}\;\;d\in (x,a)\;\;\text{or}\;\;d\in(a,x)$$
