This is the recursive formula for which I'm trying to find an asymptotic closed form by the master theorem: $$T(n)=9T(n/27)+(n \cdot \lg(n))^{1/2}$$
I started with $a=9,b=27$ and $f(n)=(n\cdot \lg n)^{1/2}$ for using the master theorem by $n^{\log_b(a)}$, and if so $n^{\log_{27}(9)}=n^{2/3}$ but I don't understand how to play with the $(n\cdot \lg n)^{1/2}$.
I think that the $(n\cdot \lg n)^{1/2}$ is bigger than $n^{2/3}$ but I'm sure I skip here on something.
I think it fits to the third case of the master theorem.