I'm trying to solve a recurrence using the master method:
$T(n) = 10*T(n/3) + n^2\log^5n$
I set up the comparison
$n^2\log^5n$ and $n^{\log_3 10}$
and reduced them to:
$\log^5 n$ and $n^{\log_3(10/9)}$
I'm struggling to understand how these functions compare asymptotically.
From one side I understand that in general polynomials grow slower than logarithms, as written in a similar question here: Sorting functions by asymptotic growth.
However, I graphed both of the functions using a graphing calculator and it appeared that the logarithmic function grew faster than the polynomial, even at large n.
Basically, is it really true that $(\log n)^{\alpha} = o(n^\beta)$ for all $\alpha,\beta>0$? If it is, how would you go about rigorously proving it one way or the other?
