I think you may have used the wrong notation. Big-Theta is "bounded below and above asymptotically".
If you're saying whether or not $c \log n > d \log (\log n)$ for some $d$ and all $n > n_0$, then it's true and here's why:
Let's take a look at the function $f(x) = x$, it grows faster than $g(x) = \log x$. You should at lease be familiar with this fact.
Consider now that $\log x$ is a monotonically increasing function, $u > v \implies \log u > \log v$
So, if $f(x) > g(x) \implies \log f(x) > \log g(x) \implies \log x > \log (\log x)$
Obviously, all the above are true for points $x > x_0$, but not for all $x$. The above logic follows when considering those appropriate $x$ values.
Now, it's on you to apply the logic to functions which are $O$ and $\Theta$ respectively.