# How to solve generally solve this equation?

I have a set of numbers $R$ with $n$ elements. For this example, let's use $R = \{2, 3, 5, 7\}$. In practice, it has about 20 to 30 elements. There's also a constant $P$. Let's use $P = 10$ for this.

I am trying to solve the following equation for $c$:

$\displaystyle\sum_{i=1}^n \left (\frac{R_i}{max(R)}\right )^c = P$

This example gets reduced too:

$\left (\frac{2}{7}\right )^c +\left (\frac{3}{7}\right )^c +\left (\frac{5}{7}\right )^c +\left (\frac{7}{7}\right )^c = 10$

I have no idea how to extract $c$ in this example, even less so in the general case. Some trial-error led to $c = -1.228079005$.

Is there a good way to compute this?

You probably cannot do better than an iterative scheme like Newton's method. Your function is $f(c)-P$, of which you are look for roots $c$. For a term like $a^c$, you have $(a^c)'=(\ln a)*a^c$. So find $f'(c)$ and iterate.
As a sanity check, make sure that $1\leq P\leq n$, otherwise this has no solution.
You won't find an algebraic approach that will work, you will have to use a numeric root finder. To get a starting value, it is probably easier to write your equation as $$\displaystyle\sum_{i=1}^n \left (\frac{\max(R)}{R_i}\right )^{-c} = P$$ and note that the term with the smallest $R_i$ will be larger (maybe much larger) than the rest so start by finding the $c$ which satisfies $\left (\frac{\max(R)}{\min (R)}\right )^{-c} = P$ and go from there.