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.