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The inequality is $1000n^3<2^n$ I'm trying to see for what value of $n$ an algorithm A who takes ($1000n^3$) steps takes less time than another algorithm B who takes($2^n$) steps.

Any idea how to do that step by step?

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marked as duplicate by Paul Sinclair, Claude Leibovici, Arnaud D., choco_addicted, Krish Nov 24 '17 at 12:24

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ Take logarithm on both sides. $\endgroup$ – Jean Marie Nov 23 '17 at 21:12
  • $\begingroup$ I don't know how in this specific case $\endgroup$ – Yoni Nov 23 '17 at 21:12
  • $\begingroup$ ... then compare the curves of $f(x)=\log(1000)+3\log(x)$ and $g(x)=x \log(2)$, which is easier that the intial curves. They intersect at a certain value $x_0$. Round it at the closest integer which is n=24. $\endgroup$ – Jean Marie Nov 23 '17 at 21:18

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