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.
 A: 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.
