would you please explain me why does below equation hold? $$\log(\log^{*}(n)) = \Omega(\log^{*}(\log (n)))$$
This means that $\log^{*}(\log (n))$ in big $n$'s has a lower growth compared to $\log(\log^{*}(n))$
On the other hand, I have read in the book that $\text{logarithmic functions}$ have the same growth(theta of each other)
Suppose we take definition of $\log^{*}$ from here where definition says "The iterated logarithm is the number of times the logarithm has to be applied before the result becomes one or less".
So if we start from $n$ and apply $\log^{*}$, suppose, this gives $k$ steps of logarithm compositions until it becomes one or less. If for same $n$ we consider as start point $\log n$, then obviously we will have $k-1$ steps of logarithms composition from $\log^{*}$. Formally, for sufficiently big $n$, we have $$\log^{*}(\log n)=\log^{*}(n)-1 > \log(\log^{*}(n))$$