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I'm not 100% sure about finding the rate of growth for logarithms, so I just want to make sure I've understood it correctly.

Is it true that the the order of growth for

$$n+\log_2(n)$$

is $\mathcal{O}(n)$ because the order of $\log_2(n)<n$?

And the order of growth for

$$n^2 + n\log_{10}(n)$$

is $\mathcal{O}(n^2)$ because the order of $\log_{10}(n^n)<n^2$?

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    $\begingroup$ Yes. It can be trivially and directly proved. $\endgroup$ Sep 17, 2016 at 14:48

1 Answer 1

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This page has a list of Time Complexities of a whole bunch of functions sorted by rate of growth.

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