Is nlogn (or more specifically nlog[base2]n) the same as: n multiplied by logn?
Yes. In a programming, it would be written something like n * log(n).
I am guessing you're talking about Big Oh notation. An interesting thing about Big Oh is that it doesn't matter what base log you're referring to. Given two bases, log_a(x) and log_b(x) will always be proportional to each other for all x. Big Oh notation ignores scalar differences between functions, so O(log_a(x)) = O(log_b(x)).
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