How to go about proving there exist some constant $c$ which would make $[ln(N)]^k \le cN$ , where k is also a constant.
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Does $\log^k (N)$ refer to $\left[\ln(N)\right]^k$ or does it refer to $\log_{k}(N)$? – user2661923 Nov 12 '21 at 22:26
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$\log^k(N)$ refers to $[ln(N)]^k$. I have also updated the post. – jnxd Nov 12 '21 at 22:27
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If you're looking for asymptotic behaviour, I'd suggest showing this with a limit: by L'Hospital's rule and induction you can show that $\lim_{N \to \infty} \frac{\ln^k(N)}{N} = 0$ for nonnegative integers $k,$ and noninteger $k$ aren't a problem because $\ln^{\lceil k \rceil}(N) > \ln^k(N).$ – Stephen Donovan Nov 12 '21 at 22:31
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Let $\displaystyle f(x) = \frac{\left[\ln(x)\right]^k}{x}$. It is routine to show that there exists a specific fixed real number $M$, which is a function of the (presumably) fixed value of $k$ such that for all $x > M, ~f'(x) < 0.$ – user2661923 Nov 12 '21 at 22:33
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Perfect. Thanks @rtybase – jnxd Nov 12 '21 at 23:00