I'd rather write $$2^{\sqrt{\log_2n}}/\sqrt n=2^{\sqrt{\log_2n}-(1/2)\log_2n}$$ and then note $$\sqrt{\log_2n}-(1/2)\log_2n\to-\infty$$ as $n\to\infty$ so the limit is, indeed, zero. This doesn't prove $2^{\sqrt{\log_2n}}\lt\sqrt n$ for all $n$, but certainly implies that inequality for all sufficiently large $n$.
Here is a more general framework for your question. Suppose you have functions $f(x)$ and $g(x)$. You compare their growth rates by looking at $$\lim_{x\to\infty}\frac{f(x)}{g(x)}$$
In your problem, $f(x)=3^x$ and $g(x)=2^x$, and this limit is $\infty$, which is what we mean when we say "$3^x$ grows faster than $2^x$."
Then you ask: why does taking the logarithm not yield the same order? (In other words, you're asking: if $f(x)$ grows faster than $g(x)$, why doesn't $\log f(x)$ grow faster than $\log g(x)$? Or, in yet other words: why don't logs preserve the asymptotic ordering?) The answer is that you must now consider $$\lim_{x\to\infty}\frac{\log f(x)}{\log g(x)}$$
And there is no nice way to relate the first limit to this limit, in general.
But in your special case when $f(x)$ and $g(x)$ are exponentials, taking the log gives first-order polynomials, which have the same growth rate: $$\lim_{x\to\infty}\frac{\log 3^x}{\log 2^x}=\lim_{x\to\infty}\frac{x\log 3}{x\log 2}=\frac{\log 3}{\log 2}$$
(Since the limit is finite and non-zero, the new functions have the same growth rate.)
EDIT
No, pure exponentials are not the only family of functions that has this property. Consider $f(x)=x\cdot 3^x$ and $g(x)=x\cdot 2^x$. These functions have different growth rates, because $$\lim_{x\to\infty}\frac{x\cdot 3^x}{x\cdot 2^x}=\infty$$
But their logs, $\log x+x\log 3$ and $\log x+x\log 2$, have the same growth rates, because
$$\lim_{x\to\infty}\frac{\log x+x\log 3}{\log x+x\log 2}=\lim_{x\to\infty}\frac{\frac{1}{x}+\log 3}{\frac{1}{x}+\log 2}=\frac{\log 3}{\log 2}$$
In fact, as this example suggests, any function that is the product of a bunch of functions all of which have growth rate at at most exponential will work: their logarithms will have the same growth rate. Take $f(x)=(\log_3 x)(x^3-4x^2+1)3^x$ and $g(x)=(\log x)(x^2)2^x$ for example. These functions have different growth rates, again, but their logarithms have the same growth rate. The logarithm turns the multiplicative structure into an additive structure dominated by a first-order polynomial.
Best Answer
For any $\alpha > 1$, we can check that
$$f(\alpha C\log C)=(\alpha-1+o(1))C\log C \qquad\text{as}\quad C\to\infty.$$
In particular, this tells that $x < \alpha C\log C$ for large $C$. On the other hand,
$$ x = C\log x = C\log(C\log x) \geq C\log C. $$
So it follows that $x \sim C\log C$ as $C\to\infty$. Throwing this to the identity $x = C\log x$, we easily check that
$$ x = C \left( \log C + \log\log C + \mathcal{O}\left( \frac{\log\log C}{\log C}\right) \right) \qquad\text{as}\quad C\to\infty. $$