While looking at the answer key for CLRS Introduction to Algorithm comparing the asymptotic behavior of $f(x) = 2^{(lg*n)}$ and $g(x) =ln(ln (n))$, it states that one can approach this by taking $lg(f(x)) = lg^*(n)$ and $lg(g(x)) = lg(ln(ln(n))$. Thus as $lg(ln(ln(n)) = \omega (lg^*(n))$ then it follows that $f(x) = \omega(g(x))$. It seems to me that they are suggesting that if $lg(f(x)) = \omega(lg(g(x)))$ then $f(x) = \omega(g(x))$. However I am having difficulty showing why this would be true.
For clarification CLRS defines $f(x) = \omega(g(x))$ if for a fixed positive constant $k$, $\exists n_o \in\mathbb N^+$ such that $f(x) > k * g(x)$ for all $x > n_o$.
lg*(n) is defined as the minimum positive integer k such that $lg^{(k)}(n) \leq 1$ where, $lg^{(k)}(n) = lg(lg(lg...(n))$ ,iterated k times ,is the iterative log function.