Asymptotic analysis - if f(n) = Ω(g(n)), how to prove ln(f(n)) = Ω(ln(g(n)))? Is the following statement true, if so, how can I prove it?
if f(n) = Ω(g(n)), is also true that ln(f(n)) = Ω(ln(g(n)))?

Since f(n) >= c*g(n), I divided the problem into two branches:
If f(n) = g(n), it's also true that ln(x) = ln(x) and the statement it's true.
If f(n) > g(n), then the value for ln is greater, if the argument is greater:
x > y; ln(x) > ln(y)

Is there something wrong with my reasoning?

This is false in general without some further conditions.
For example take $f(n) = n$ and $g(n) = 2n$.  Then $f(n) \geq \frac{1}{2} g(n)$ on the domain $n \geq 1$ but $\log f(n) \not\geq c \log g(n)$ on the domain $n \geq 1$ for any $c > 0$.  In this case it is only true that $\log f(n) = \Omega(\log g(n))$ as $n \to \infty$.
For another example take $f(n) = \sin(n\pi/6) + 2$ and $g(n) = \sin(n\pi/6) + 3$.  Then $f(n) \geq \frac{1}{2} g(n)$ for all $n$ but $\log f(n) \not\geq c \log g(n)$ for any $c > 0$ on any domain that includes the points $n=9,21,33,45,\ldots$
The statement is true if $g(n) \to \infty$ as $n \to \infty$ and $f(n) = \Omega(g(n))$ as $n \to \infty$.  For then we can find a $c > 0$ such that
$$
f(n) \geq c\,g(n)
$$
for all $n$ sufficiently large, from which
$$
\log f(n) \geq \log c + \log g(n) \geq \frac{1}{2}\log g(n)
$$
for $n$ large enough.  Thus $\log f(n) = \Omega(\log g(n))$ as $n \to \infty$.
