Since you mentioned the comparison test... First notice that
$$2^{-n/2}\sqrt{\log{(2^n)}} = \frac{\sqrt{n\log{(2)}}}{\sqrt{2^n}} = \sqrt{\frac{n\log(2)}{2^n}},$$
so
$$\sum_{n=1}^\infty 2^{-n/2}\sqrt{\log{(2^n)}} = \sqrt{\log(2)}\sum_{n=1}^\infty \sqrt{\frac{n}{2^n}}.$$
Next, notice that $2^{n/2}\geq n$ for any $n\geq 4$. Hence:
$$\sum_{n=4}^\infty 2^{-n/2}\sqrt{\log{(2^n)}} = \sqrt{\log(2)}\sum_{n=4}^\infty \sqrt{\frac{n}{2^n}} \leq \sqrt{\log(2)}\sum_{n=4}^\infty \sqrt{\frac{2^{n/2}}{2^{n}}} = \sqrt{\log(2)}\sum_{n=4}^\infty \sqrt{\frac{1}{2^{n/2}}}=\sqrt{\log(2)}\sum_{n=4}^\infty \frac{1}{2^{n/4}},$$
and the last series is clearly convergent! Thus, the original series is convergent as well.