Master Method: $T(n) = 10T\Big(\frac{n}{2}\Big) + \frac{n^4}{\log(n)}$

Question

I'm having a hard time trying to understand how to solve this recurrence relation using the Master Method:

$$T(n) = 10T\Big(\frac{n}{2}\Big) + \frac{n^4}{\log(n)}$$

First, we have:

$a = 10,\ b = 2$ so we have $n^{\log_2^{10}} = n^{\lg10}$

Now we need to find how $f(n)$ compares to $n^{\lg10}$.

And here I'm stuck. What do I have to pay attention to in order to know whether I'm dealing with Case 1 (and look for $O(n^{{\lg10} - \varepsilon}))$ or Case 3 (and look for $\Omega(n^{{\lg10} + \varepsilon})) $ of the method?

My attempt:

From what I know $\log(n) < n^\varepsilon\quad \forall \varepsilon > 0$. So

$$\frac{1}{\log(n)} > \frac{1}{n^\varepsilon}$$ $$\frac{n^4}{\log(n)} > \frac{n^4}{n^\varepsilon}$$

$$\implies \frac{n^4}{\log(n)} = \Omega( n^{4-\varepsilon})$$

but I have no idea how to relate this (assuming no mistakes) with one of the two cases. In Case 1 we deal with $O(\cdot)$ and not $\Omega(\cdot)$ while in the 3rd Case the form of the exponents doesn't seem to match.

Answer 1

In order to apply Case 3, you have to show that $\frac{n^4}{\log n} = \Omega(n^c)$ for some $c > \log_2 10$, as well as the regularity condition (which holds for functions of the form $n^\alpha \log^\beta n$).

You take it from here. Hint: $4 > \log_2 10$.