Big-O notaion for $2n^4 + \log_2n^8$

Question

I need to find the best big-O for:

$$3n^4 + \log_2n^8$$ So I said: $$3n^4 + \log_2n^8 = 3n^4 + 8\log_2n \leq 3n^4 + 8n$$ Therefore, the given function is $O(n^4)$ in the best case.

Is this correct?

a small editing advice: if you a / to $log_2$ then it looks like $\log_2$ — user190080, Jul 19 '15 at 21:37

man_in_green_shirt · Accepted Answer · 2015-07-19T21:56:03.037

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The final answer is correct. Note, however, that if the $3n^4$ term was missing, your answer would be $\mathcal{O}(\log_2n)$, not $\mathcal{O}(n)$.

edited Jul 19 '15 at 21:56

answered Jul 19 '15 at 21:40

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I think we can say that it would then be $O(\log_2n)$, right? (not $O(\log_2n^8$)). – Julia Jul 19 '15 at 21:44
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Yes, you're right, sorry for the oversight – man_in_green_shirt Jul 19 '15 at 21:55

1 Answers1