#### Answer

$\ln x$ grows faster than $\ln(\ln x)$.

#### Work Step by Step

We will find the limit $\lim_{x\to\infty}\frac{\ln x}{\ln(\ln x)}.$
1) If it is equal to zero then $\ln x$ grows slower than $\ln(\ln x)$;
2) If it is equal to $\infty$ then $\ln x$ grows faster than $\ln(\ln x)$;
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"
$$\lim_{x\to\infty}\frac{\ln x}{\ln (\ln x)}=\left[\frac{\ln\infty}{\ln(\ln\infty)}\right]=\left[\frac{\infty}{\ln\infty}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(\ln x)'}{(\ln (\ln x))'}=\lim_{x\to\infty}\frac{(\ln x)'}{\frac{1}{\ln x}(\ln x)'}=\lim_{x\to\infty}\frac{1}{\frac{1}{\ln x}}=\lim_{x\to\infty}\ln x=\left[\ln\infty\right]=\infty,$$
and thus $\ln x$ grows faster than $\ln(\ln x)$.