# Solve equation with unknown in exponents

This is in continuation of this but not related to it completely.

I am interested in finding a solution to the equation: $m' = m - \sum \limits_{j=1}^{m} (1 - d_{O_j}/n)^k$. where $m,m',n$ and $d_{O_j}$ for $j \in {1,2,...m}$ are known and are positive. The only unknown is $k$. How can I approach this problem?

The presence of summation and exponent implies I cannot apply the basic approach of taking log on both sides. Any other approaches will be helpful.

Making it more compact, it seems to me that the equation you consider can write $$F(k)=\sum_{j=1}^m a_j^k-b=0$$ where all terms are known except $k$ (which I suppose to be a real number).
I really do not know if any analytical method could apply to the problem. If I am right, then only numerical methods could be used (such as Newton as the simplest) considering $k$ as a continuous variable.
Minimization of $$\Phi(k)=\Big( \sum_{j=1}^m a_j^k-b\Big)^2$$ could be an alternate solution for solving the problem.
If the case where all $a_j$ and $b$ are positive, function $F(k)$ could be written as $$F(k)=\sum_{j=1}^m e^{k \log(a_j)}-b=0$$ and then solving for $k$ $$G(k)=\log\Big(\sum_{j=1}^m e^{k \log(a_j)}\Big)-\log(b)=0$$ could be quite efficient since the function would be quite smooth and not very far from linearity.
A first order Taylor expansion should provide a reasonable approximation $$k_0=\frac{m \Big(\log (b)-\log (m)\Big)}{\sum _{i=1}^m \log (a_i)}$$ for starting Newton iterations.