For a simple summation: $$ Z = \sum_{x=1}^X x^n $$
Solve for n. I have googled unsuccessfully and I do not remember this.
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$\begingroup$You won't find a nice algebraic solution, so to get an accurate answer you will need to use a numeric root finder. To get started, you can approximate the sum with an integral. $$\int_0^{N-1}x^n\; dx \le \sum_{n=1}^N x^n\le \int_1^N x^n\; dx\\ \frac {(N-1)^{n+1}}{n+1}\le \sum_{n=1}^N x^n\le \frac {N^{n+1}-1}{n+1}\\$$ so you can start from $n \approx \frac {\log Z}{\log N}-1$ If you want a little closer starting approximation, you can start with that and iterate $n_i= \frac {\log Z-\log(n_{i-1}+1)}{\log N}-1$ to convergence.
$\endgroup$ 7 $\begingroup$This is an addendum to Ross' answer, which on one hand assumes a geometric series (different from the OP's question) but on the other hand is accepted anyway. Ross' ansatz, whatever meant, can be improved
If indeed the geometric series (where the index goes into the exponent of the sum-term) were intended then there is a much simpler answer: $$ \begin{eqnarray} Z&=& \sum_{k=1}^n x^k = x{x^n-1\over x-1} \\ Z \cdot {x-1\over x} &=& x^n -1 \\ Z \cdot {x-1\over x}+1 &=& x^n \\ n &=& \log_x \left( Z \cdot {x-1\over x}+1 \right) \end{eqnarray}$$
$\endgroup$ $\begingroup$You cannot solve that which is neither a problem nor an equation :). What you likely mean ism how do i express the sum $$ S = \sum_{n=1}^N x^n $$ in terms of $N$, its upper limit.
To do that, what would happen if you were to compute $S - S \cdot x$? Can you derive the formula for $S$ from here?
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