question about differentiation on logarithmatic functions I am a little confused as to why sometimes they will treat $x$ in $\ln x$ as $\ln ab$ and other times its treated as $\ln x$.... To be more clear, here is an example:
$2\ln(3x^2-1)$  and $\ln2x$. Now, in $2\ln(3x^2-1)$ the $3x^2-1$ becomes $x$ and is written as $2\cdot1/x\cdot x'$.
My question is why? because this could have been treated as $\ln3x^2/\ln1$.
On the other hand, $\ln2x$ they solve as $\ln2+\ln x$ then differentiate it.... So my question is: When do they choose to apply the log rules and when not to? Will not applying the rule give you the wrong answer? In the case of $2\ln(3x^2-1)$ could they have done it differently as oppose to treating $3x^2-1$ as $x$.
 A: $\ln(3x^2-1)$ is certainly not the same as $\ln(3x^2)/(\ln 1)$.
Correct: $\ln\dfrac ab=(\ln a) - (\ln b)$
Incorrect: $\ln(a-b) = \dfrac{\ln a}{\ln b}$
A: 
Now, in $2\ln(3x^2-1)$ the $3x^2-1$ becomes $x$ and is written as $2\cdot1/x\cdot x'$.

I'll assume you mean it's treated as another variable (not $x$), perhaps $u$, and you differentiate $2\ln u$ to obtain $2\cdot\frac1uu'$ (which is then $2\cdot\frac1{3x^2-1}6x$.
The answer is that it doesn't matter. You can do either method in either case. But whereas $\ln 2x$ is easy to split up into $\ln2+\ln x$, $\ln(3x^2-1)$ is not as easy to split up — see Michael Hardy's answer — so they don't split it up. (But if you'd find a way to split it up, you could.)
A: I hope that the following comments settle some of your questions. Unfortunately, it is not clear what notation you use for the Chain Rule.
1) Suppose that $H(x)=\ln(5x)$. We want to find $H'(x)$.  Note that $\ln(5x)$ is "function of a function of $x$." Specifically, $\ln(5x)=f(g(x)$, where $g(x)=5x$ and $f(u)=\ln(u)$.
Therefore, by the Chain Rule, we have $H'(x)=g'(x)f'(g(x))$.  Note that $g'(x)=5$ and $f'(u)=\dfrac{1}{u}$.  So $f'(g(x))=\dfrac{1}{5x}$.
Now put things together. We have $H'(x)=(5)\dfrac{1}{5x}$.  This simplifies to $\dfrac{1}{x}$.
In this case, we could have avoided the Chain Rule, by noting that $H(x)=\ln(5x)=\ln(5)+\ln(x)$.  Differentiate, using the fact that $\ln(5)$ is a constant and therefore its derivative is $0$. We get $H'(x)=\dfrac{1}{x}$.
2) Let $H(x)=\ln(3x^2-1)$.  We want $H'(x)$.  Note that $H(x)=f(g(x))$ where $g(x)=3x^2-1$ and $f(u)=\ln(u)$. Use the Chain Rule like before. We have $g'(x)=6x$ and $f'(u)=\dfrac{1}{u}$. Putting things together like before, we get $H'(x)=\dfrac{6x}{3x^2-1}$. No useful simplification is available.
Important note: It looks as if your post asks why we can't treat $\ln(3x^2-1)$ as $\dfrac{\ln(3x^2)}{\ln(1)}$.
The reason that we can't is that $\ln(3x^2-1)$ is not equal to  $\dfrac{\ln(3x^2)}{\ln(1)}$.  The logarithm of a product is a sum of logarithms. the logarithm of a quotient is a difference of logarithms. So for example, $\ln((x^2+1)(x^2+5))=\ln(x^2+1)+\ln(x^2+5)$, and $\ln(7/5)=\ln(7)-\ln(5)$.
But the logarithm of a sum can't be "simplified" in a pleasant way, and neither can the logarithm of a difference.  You can check for yourself, with a calculator, that $\ln(7-2)$ is not equal to $\dfrac{\ln(7)}{\ln(2)}.$
If you use the FALSE "rules" $\ln(a+b)=\ln(a)+\ln(b)$ or $\ln(a-b)=\dfrac{\ln(a)}{\ln(b)}$, it will cause you grief over and over. 
