How do I take the $\log$ of a inequality of the form: $0$0<x<1$ taking the $\ln$ would give $\ln0<\ln x< \ln 1 \implies \text{undefined} < \ln x <0 $. 
If $x=1/2$ I could use $1/4<1/2<1$ and would be ok. But I only know that $x>0$. This means I can't take the $\log$ of the inequality mentioned?  
 A: It would be fruitful to consider the behavior of $\ln(x)$ as $x\to 0$. A graph helps if the notion of this sort of thing (formally called in calculus as a "limit") is new to you, but for now just imagine testing $\ln(x)$ for smaller and smaller values of $x$: $1/2, 1/10, 1/100, 1/1000$, and so on, getting closer and closer to $0$. You'll notice that $\ln(x)$ begins to blow up to $-\infty$.
The graph of $\ln(x)$, to make this point clear:

If you consider the behavior of the function, you can sort of take the logarithm of the inequality, where if the actual value is undefined, you can try the limit and see if that works. Sure, $\ln(0)$ is undefined - but as $\ln(x)$ approaches $0$, $\ln(x)$ becomes more and more negative. In the formal calculus notation, we would say
$$\lim_{x \to 0^+} \ln(x) = -\infty$$
where the $+$ is meant to indicate "approaching from the right" (you can see why this is noted since there really isn't a "from the left" in this scenario).
(Footnote: This idea of translating the undefined values into limits at that point doesn't work for every scenario, but as you become more sophisicated mathematically and work more examples, you'll get a feel of when you can do this when.)
In any event, we would have
$$0<x<1 \implies -\infty < \ln(x) < \ln(1) = 0$$
Of course, you can drop the $-\infty$ bit since $\ln(x) < 0$ implies that too.
A: I'm not that sure, but when applying a function that approaches to positive infinity, you simply apply the function on all sides of the equality/inequality, without ay issues. Undefines & infinities are all consistent.
When a domain (a pre-image) in consideration is within one that produces a range (image) that falls within negative, it also shouldn't be an issue.
Say you multiply a constant of 5 on all sides ($f(x) = 5x$), then you don't have to invert the relational operators. Multiply by -5 ($f(x) = -5x$),  & you invert them.

MORE:
There is a hitch, if you will. What about multiplying by c, thus $f(x) = cx$ ? Then you don't invert it yet. You only apply the inversion as the actual numerical value of the variable c is determined. You determine it to be -4, then invert relational operators.
