How can I re-write $2^{\sqrt{\log n}}$ as $n^?$

Question

How can I re-write $2^{\sqrt{\log n}}$ as $n^?$

I tried $2^{\log n^{0.5}}$ then $2^{0.5\log n}$, then $n^{0.5 \cdot 1} = \sqrt{n}$.

But it seems to be smaller than $\sqrt{n}$ in the answers sheet.

Answer 1

$2^{\sqrt{\log(n)}}$ does not equal $n^a$ for any constant $a$.

The calculation you sketch is wrong because you fail to distinguish between $2^{(\log n)^{1/2}}$ and $2^{\log(n^{1/2})}$ at the beginning.

If you're looking for asymptotics for $n\to\infty$ what you can say is $$ 2^{\sqrt{\log n}} = 2^{\left(\frac{\log n}{\sqrt{\log n}}\right)} = n ^ {1/\sqrt{\log n}} $$ where the exponent $1/\sqrt{\log n}$ goes to $0$, so $2^{\sqrt{\log n}}$ is $O(n^\varepsilon)$ for every $\varepsilon>0$.

On the other hand, since $\sqrt x$ grows faster than $\log x$, we have that $2^\sqrt{x}$ grows faster than $x$, and therefore $2^{\sqrt{\log n}}$ is $\Omega(\log n)$.