Master theorem recurrence relation Consider I have the following recurrence

$$T(n) = 10T(n/3) + \Theta(n^2\log^5 n)\,.$$

Now, by the master theorem, if we evaluate $n^{\log_{b}{a}}$, we get $n^{\log_{b}{a}} = n^{\log_{3}{10}} = n^{2.095}$. Now, can anyone please explain me, which of the three cases apply? If you plot these functions you can see that $n^2\log^5 n$ clearly dominates. So, the answer should be $\Theta(n^2\log^5n)$.

However, if you try plugging in values in this calculator, you see that the answer is $\Theta(n^{\log_{3}{10}})$. Does anyone have a formal reason? I am still confused as to which of these conditions apply.
Plotting isn't foolproof: for any $k,\varepsilon&gt;0$, we have $\log^kn = o(n^\varepsilon)$ (here, I've cancelled the common term of&nbsp;$n^2$). However, you can need to take $n$ really big to see it when $\varepsilon$&nbsp;is small (e.g., $0.095$) and $k$&nbsp;isn't so small (e.g., $5$).  Here's the plot for $0\leq n\leq 10^{131}$, again with the $n^2$&nbsp;term cancelled, and it looks quite different from your plot up to $n=10^5\,$!

Indeed, this question may even have been designed to catch you out if you try to compare the asymptotic growth by plotting instead of by understanding.