Is $n\log n = \Omega (n^{log_4(3) + \varepsilon})$?

Question

I'm reading CLRS (Introduction to Algorithms). The following in an example of how to use the master theorem to determine the complexity of a recurrence.

$$T(n) = 3T\left(\frac{n}{4}\right) + n\log .$$

with $a = 3, b = 4.$

$$f(n) = n\log n.$$

$$n^{\log_b(a)} = n(\log_4(3)) = O(n^{0.793})$$

The following statement is what I don't understand.

Since $$f(n) = \Omega(n^{log_4(3) + \epsilon})$$ where $\varepsilon ≈ 0.2$, case $3$ applies.

Where in the world did that (^) come from?

Answer 1

$3<4$, so $\log_4(3)<1$, so there's some small $\epsilon$ such that $\log_4(3)+\epsilon<1$. This implies $n^{\log_4(3)+\epsilon} < n^1 < n\log n$, for sufficiently large values of $n$.