# Solve an equation with arbitrary exponents

I reduced a system of non-linear equations to a single equation for $c$:

$$\left(\frac{x-c}{y-c}\right)^{w - u} = \left(\frac{x-c}{z-c}\right)^{v-u}$$

I need to solve this equation for $c \notin \{ x, y, z \}$ knowing $x, y, z, u, v, w \in \mathbb{R}$ where $x \neq y, x \neq z, y \neq z$ and $u \neq v, u \neq w, v \neq w$.

Is there any way to solve this equation directly? I would like to avoid iterative algorithms as part of this problem. If there is no direct way to solve the equation, which algorithm would you suggest?

• Just to be clear, we are solving for $c$, right? – Simply Beautiful Art Feb 19 '17 at 23:26
• @SimplyBeautifulArt Yes, I hope I am missing something obvious, I am stuck at this for hours. – just.kidding Feb 19 '17 at 23:27
• Nah, it's definitely not solvable algebraically. – Simply Beautiful Art Feb 19 '17 at 23:27
• It is algebraic if $(w-u)/(v-u)$ is rational, but that's not going to help you: the polynomial in $c$ will generally not be solvable in terms of radicals. If you don't want to use iterative algorithms such as Newton's method, you might have power series solutions in some cases. – Robert Israel Feb 19 '17 at 23:34

If we multiply all the denominators away, we get

$$(x-c)^{w-u}(z-c)^{v-u}=(x-c)^{v-u}(y-c)^{w-u}$$

For simplicity, let $w-u=a$ and $v-u=b$.

$$(x-c)^a(z-c)^b=(x-c)^b(y-c)^a$$

Before I tackle $a,b\in\mathbb R$, let's look at $a,b\in\mathbb N$.

Unfortunately, for $a,b\ge5$, there exists no closed form solution of $c$ in terms of radicals for general $x,y,z\in\mathbb R$ due to the Abel-Ruffini theorem. It follows that we can't expect a closed form for $a,b\in\mathbb R$ either...

To be suggestive, Newton's method should work nicely. Here, we have

$$c_{n+1}=c_n-\frac{(x-c_n)^{a-b}(z-c_n)^b-(y-c_n)^a}{\left(\frac{a-b}{x-c_n}+\frac b{x-c_n}\right)(x-c_n)^{a-b}(z-c_n)^b-a(y-c_n)^{a-1}}$$