Master theorem: When a $f(n)$ is smaller or larger than $n^{\log_b a}$by less than a polynomial factor
$ T(n) = 25T(n/5) + 20(n\log n)^{1.99} $


Note that $(\log n)^m=O(n^\epsilon)$ for all $m,\epsilon\in \Bbb R^+$. In plain words, all polynomials of $\log n$ grows slower asymptotically than all polynomials of $n$. Here we abuse "polynomials" to include functions like $x\to x^k$ for all $k\in\Bbb R^+$ such as $x\to x^{0.618}$ or $x\to x^{2019.46}$. This abuse of the term "polynomial" is common when we are talking about the asymptotic behavior of functions. 

In particular, $(\log n)^{1.99}=O(n^{0.005})$.

Since $(n\log n)^{1.99}=n^{1.99}(\log n)^{1.99}= O(n^{1.995})=O(n^{log_525-0.005})$, you can apply the case 1 of Master theorem. $T(n)=\Theta(n^2)$.



If $f(n)$ is indeed smaller or larger than $n^{\log_b a}$by less than a polynomial factor, then we are running into the cases strictly between case 1 and case 2 or between case 2 and case 3. The master theorem you were reviewing cannot be applied any more. 

You may look into this answer that handles some of those situations.



Exercise. Show that $(\log n)^m=O(n^\epsilon)$ for all $m,\epsilon\in \Bbb R^+$. (For example, $(\log n)^{10000000}=O(n^{0.000001}).$)