Answer
$\ln\sqrt{x}$ grows slower than $\ln^2{x}$.
Work Step by Step
We will find the limit $\lim_{x\to\infty}\frac{\ln\sqrt{x}}{\ln^2 x}.$
1) If it is equal to zero then $\ln\sqrt{x}$ grows slower than $\ln^2x$;
2) If it is equal to $\infty$ then $\ln\sqrt{x}$ grows faster than $\ln^2 x$;
3) If it is equal to some non zero constant then their growth rates are comparable.
"LR" will stand for "Apply L'Hopital's rule". Also, note that $\ln\sqrt{x}=\ln x^{1/2}=\frac{1}{2}\ln x$ so we have
$$\lim_{x\to\infty}\frac{\ln\sqrt{x}}{\ln^2 x}=\lim_{x\to\infty}\frac{\frac{1}{2}\ln x}{\ln^2 x}=\lim_{x\to\infty}\frac{1}{2\ln x}=\left[\frac{1}{2\ln\infty}\right]=\left[\frac{1}{\infty}\right]=0, $$
and thus $\ln\sqrt{x}$ grows slower than $\ln^2{x}$.
