Question 894297
Can you take the natural logarithm of a negative number?
{{{e^(p)=ValueLessThanZero}}}
NO!  


This means that {{{2x-27>=0}}}
{{{highlight(x>=27/2)}}}
That is the necessary domain for your f(x).  Note that this is also more than sufficient for the {{{x>=3}}}  restriction of the numerator.