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This is how I would simplify the first function:

$2^{\sqrt{2\log(n)}} = 2^{\sqrt{2} \sqrt{\log(n)}} = 2^{\sqrt{\log(n)} \sqrt{2}} = (2^{\sqrt{\log(n)}})^{\sqrt{2}} = (2^{(\log(n))^{\frac{1}{2}}})^{\sqrt{2}} = n^{\frac{\sqrt{2}}{2}}$

Are my power rule applications correct?

Thank you in advance!

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    Nope: $$ 2^{\left((\log n)^{1/2}\right)}\neq \left(2^{\log n}\right)^{1/2} $$ since exponentiation **is not** associative. – 2017-01-31

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Exponentiation is not associative, hence in general $a^{\left(b^c\right)}\neq \left(a^b\right)^c$. However, if $n>e$, $$ 2^{\sqrt{2\log n}} = \exp\left(\sqrt{2}\log(2)\sqrt{\log n}\right) \color{red}{\leq} \exp\left(\sqrt{\log n}\right) \leq \exp\left(\log n\right)=n $$ since: $$ \log(2) = \int_{0}^{1}\frac{dt}{1+t}\stackrel{CS}{\color{red}{\leq}}\sqrt{\int_{0}^{1}dt\int_{0}^{1}\frac{dt}{(1+t)^2}} = \frac{1}{\sqrt{2}}.$$

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    what is 'exp', 'CS'? thanks. – 2017-01-31
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    @miniparser: $\exp(z)$ stands for $e^z$, and $CS$ for the Cauchy-Schwarz inequality. – 2017-01-31
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    @JackD'Aurizio That makes a neat bound on $\log n$ ($\sqrt{n-1}$ for $n \geq 1$) that I haven't seen before. Thanks for the excellent answer. – 2017-02-01