Is $n^3$ an asymtotically tight bound of $(n^{2.99}).(\log_2n)$? How? [duplicate]

Is $$n^3$$ an asymtotically tight bound of $$(n^{2.99}).(\log_2n)$$? If so then how?

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No. $$n^{2.999}\log_2n$$ is tighter, for example.
$$g(n)$$ is an asymptotically tight bound of $$f(n)$$ if $$f(n) = O(g(n))$$, but $$f(n) ≠ o(g(n))$$. In this case, $$n^{2.99} \log n = o (n^3)$$. For very large n, $$n^{0.01}$$ grows faster than $$\log n$$.
• I am afraid your definition of "an asymptotically tight bound" is not (equivalent to) the most common one. I would say "g(n) is an asymptotically tight bound of f(n) if and only if $f(n) = \Theta(g(n))$. – Apass.Jack Oct 10 '18 at 4:08
• @Apass.Jack That is almost the same. Yuval's definition admits some funky $f$. – Raphael Oct 10 '18 at 6:13
• @Raphael, yes, I agree these two definitions are almost the same. However, you know they are NOT the same. I just wanted to point out that it is better to simply say, "$g(n)$ is an asymptotically tight bound of $f(n)$ implies $f(n) = O(g(n))$, but $f(n) ≠ o(g(n))$". Where is Yuval's definition? – Apass.Jack Oct 10 '18 at 7:06