$2^{O(\log \log n)} = O(\log n)$ prove or disprove

Question

I need to prove or disprove this: $$2^{O(\log\log n)} = O(\log n)$$

I haven't found anything like that through search. I would like to have some help. Thanks.

Answer 1

Note that the function $10 \log_2 \log n$ is $O(\log \log n)$. But, $2^{10 \log_2 \log n} = 2^{\log_2 ((\log n)^{10})} = (\log n)^{10}$, which isn't $O(\log n)$.

Answer 2

Just transform to $\log_2$ :

$$2^{c \log \log n}=\left ( 2^{\frac{\log_2(\log n)}{\log_2 e}} \right)^c=\left (\log n \right )^{\frac{c}{\log_2 e}}$$