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I am trying to calculate the $k$ value in this equation:

$\dfrac1{n^c} \le \left(1 - \dfrac2{n(n-1)} \right)^k$

by using the logarithm, I am getting for $k$:

$\log_{1- 2/n(n-1)} n^{-c} \le k$

is that correct? And ist there any way to simplify the result of $k$?

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If you take logs, $\dfrac1{n^c} \le \left(1 - \dfrac2{n(n-1)} \right)^k $ becomes $-c\log(n) \le k \log\left(1 - \dfrac2{n(n-1)} \right) $.

Since $\log\left(1 - \dfrac2{n(n-1)} \right) < 0$, dividing by it reverses the sign of the inequality, so we get $\frac{-c\log(n)}{\log\left(1 - \dfrac2{n(n-1)} \right)} \ge k $.

If $n$ is large compared with $k$, the right side is about $1 - \dfrac{2k}{n(n-1)} $, so we get, as an approximation, $\dfrac1{n^c} \le 1 - \dfrac{2k}{n(n-1)} $ or $k \le \frac12 n(n-1)(1-n^{-c}) $.

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