Big-O notation, logs vs exp

196 Views Asked by At

I have simplified these functions to the best of my ability. I know logs are less than exp, but are all exponential functions greater than logs, even if the power is less than 1?

$$n^{1.01} = \Omega(n(\log n)^2) \Rightarrow n^{.01} = \Omega((\log n)^2))$$