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As the topic says, I need to simplify:

$$\ln |x-x^2| - \ln |x-1| $$

I don't know how to approach the problem at all. I'm not asking for the answer, but something to maybe get me going.

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up vote 1 down vote accepted

Remember that logarithms have some rules associated with them that help you simplify problems. The most common are: $\ln(ab) = \ln a + \ln b$ and $\ln(a/b) = \ln a - \ln b$. The second one is the key to your problem. If you have the difference of logs, say $\ln a - \ln b$ then you can simplify it to be the log of a quotient: $\ln(a/b)$. If you have a quotient of polynomials, then you should be trying to factorise and eliminate common factors. For example:

$$ \frac{x^2-x}{x-1} = \frac{x(x-1)}{x-1} = \ ? $$

You want to simplify $\ln |x^2-x| - \ln |x-1|,$ so apply the two steps that I suggest.

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So I act like there is no absolute value expression here? How come that is possible, if so? I think it's the absolute value expression that is making this hard for me. For example, $\ln{\left(\frac{|x(x-1)|}{|x-1|}\right)}$ - it's not possible to cancel the $x-1$ out in this case, or am I wrong? Edit: Yes, you're (I'm) wrong. $|ab| = |a||b|$ ... – matfor Sep 12 '12 at 17:59
Too slow to edit my last comment. Adding another one (against SE policy?). Since logarithm functions only accepts positive numbers, we don't have to set up cases (when positive, when negative) and solve for each? – matfor Sep 12 '12 at 18:08
Strictly speaking, logarithms can accept positive, negative and even complex values. My arguments $a$ and $b$ are arbitrary. Just because I didn't include a modulus doesn't mean you can't. Replace $a$ by $|x|$ and $b$ by $|y|$ if it makes you happy. You can cancel: $$\frac{|x^2 - x|}{|x-1|} = \frac{|x(x-1)|}{|x-1|} = \frac{|x|\cdot |x-1|}{|x-1|} = |x| \, . $$ – Fly by Night Sep 12 '12 at 18:30
@matfor: Consider a complex number $z \in \mathbb{C}$. We can rewrite $z = Re^{i(\theta + 2\pi k)}$. Then $$ \ln z = \ln(Re^{i(\theta + 2\pi k)}) = \ln R + \ln (e^{i(\theta + 2\pi k)}) \, . $$ It follows that $\ln z = \ln |z| + i \, \text{Arg} \, z$ where $\text{Arg} \, z = \{ \theta + 2\pi k: k \in \mathbb{Z} \}.$ So we see that $\ln$ is a multivalued function: – Fly by Night Sep 12 '12 at 18:38
You don't have to divide: $\log(|x^2-x|)=\log(|x||x-1|)=\log(|x|)+\log(|x-1|)=\log(|x|)+\log(|1-x|)$. – André Nicolas Sep 12 '12 at 20:46

Hint: can you factor $x-x^2$ then use the laws of logs?

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