# For what value of n this inequality is true and how to find it step by step? [duplicate]

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?

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

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

• Take logarithm on both sides. – Jean Marie Nov 23 '17 at 21:12
• I don't know how in this specific case – Yoni Nov 23 '17 at 21:12
• ... 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. – Jean Marie Nov 23 '17 at 21:18