Solve a recurrence using the master theorem
$f(n) = (n\cdot \lg n)^{1/2}$ and $n^{\log_b a}=n^{2/3}$, thus
$f(n) = O(n^{\log_b a})$ and even $f(n) = O(n^{\log_b a - \epsilon})$ for $\epsilon &lt; 1/6$.

Why? because
$$\lim_{n\to\infty} \frac{f(n)}{n^{\log_b a - \epsilon}} = \lim_{n\to\infty}\frac{n^{1/2}\lg^{1/2}n}{n^{2/3-\epsilon}} =  \lim_{n\to\infty} \frac{\lg^{1/2}n}{n^{1/6-\epsilon}} =0 \quad\text{for }\epsilon&lt; 1/6$$

Thus case 1 of the Master theorem should apply, and $T(n) = \Theta(n^{2/3})$.