How can I solve this problem with master theorem. Giving asymptotic upper and lower bounds
If
$T(n)=4T(n/3)+ n log(n)$
a=4 b=3 k=1
for the formula $aT(n/b)+n^k log^b(n)$
if $a>b^k$ then $T(n)=\theta(n^{log_b^a})$ which is
then does this mean the upper and lower bound is
$=\theta(n^{log_3^{4}})$
Would this be correct I am not good at these types of problem they are.. too abstract.
The first case you need to analyze is: Is $n\log(n)=O(n^{c})$ where $c<\log_{3}(4)$? $\log_{3}(4)$ is a little bigger than $1$, so you can take $c>1$... say $c=1+\epsilon$ for a fixed small $\epsilon$. If $n\log(n)=O(n^{1+\epsilon})$--which it is, I'll leave it to you to justify that step--then by the result is that $T(n)=\Theta(n^{\log_{3}(4)})$.
This all comes from https://en.wikipedia.org/wiki/Master_theorem#Case_1