Determine the big $O$ notation

Question

For $\epsilon>0$, why the following holds with $x$ larger enough $$O(x^{-\epsilon/2}\log x)=O(2^{-\log^{1/2}x})$$

Answer 1

All you need is $\log(x) = o(x^c)$ for any $c > 0$.

Taking logs of the question, this gives $-c \log(x)$ on the left where we can choose $c = \epsilon/4$ and $-d \log^{1/2}(x)$ on the right for some $d > 0$ and this makes it clear since $\log(x) \to \infty$.