# Helping turning this into sum/difference logarithm?

Hey guys so i'm trying to turn this equation into it's sum/difference logarithm. However, the part that messes me up is turning the bottom of the fraction $$\log\left(\frac{x^2 +2x+1}{x^2 -3x +2}\right)^2\;.$$ I think it will turn into this: $$4\log(x+1) -2\big(\log(x-1) +\log(x-2)\big)\;,$$ but I don't want it to be $(x-1)^2$ times $(x-2)^2$. How do I solve it so it's $\big((x-1)(x-2)\big)^2$?

-
Since the question and both answers posted so far neglect this aspect, let me mention that nearly every $\log(x-a)$ on this page should be replaced by $\log|x-a|$. For example, the first logarithm is not $4\log(x+1)-2(\log(x-1)+\log(x-2))$ (nor is it $4\log(x+1)-2\log((x-1)(x-2))$). – Did Nov 18 '12 at 10:30

I assume you want to simplify $$\log \left( \left( \dfrac{x^2+2x+1}{x^2 - 3x+2} \right)^2\right)$$ First recall the following properties of logarithm. \begin{align} 1. & \log (a^m) = m \log (a)\\ 2. & \log \left( \dfrac{a}b\right) = \log a - \log b\\ 3. & \log \left( ab\right) = \log a + \log b \end{align} Using the first property, we get that $$\log \left( \left( \dfrac{x^2+2x+1}{x^2 - 3x+2} \right)^2\right) = 2 \log \left( \dfrac{x^2+2x+1}{x^2 - 3x+2}\right)$$ Now using the second property, we get that $$\log \left( \dfrac{x^2+2x+1}{x^2 - 3x+2}\right) = \log (x^2+2x+1) - \log (x^2-3x+2)$$ Next note that $(x^2 + 2x + 1) = (x+1)^2$ and $x^2 - 3x + 2 = (x-1)(x-2)$. Hence, we have that $$\log (x^2+2x+1) = \log ((x+1)^2) = 2 \log(x+1)$$ using property $(1)$. Similarly, we have that $$\log (x^2-3x+2) = \log ((x-2)(x-1)) = \log(x-2) + \log(x-1)$$ using property $(3)$. Putting all these together, we get what you want.

-
Marvis, so does that mean that for the denominator it should be 2log(x-2) + 2log(x-1)? Because the entire thing is being powered to a power of 2. – Kelsey Abreu Nov 18 '12 at 10:00
@KelseyAbreu I have taken into account the entire power $2$ when I wrote "Using the first property, we get that $$\log \left( \left( \dfrac{x^2+2x+1}{x^2 - 3x+2} \right)^2\right) = 2 \log \left( \dfrac{x^2+2x+1}{x^2 - 3x+2}\right)"$$ – user17762 Nov 18 '12 at 10:01

$$\log((x^2 +2x+1) / (x^2 -3x +2))^2$$ $$\log((x+1)^2 / (x^2 -2x-(x-2))^2$$

$$\log((x+1)^2 / (x(x-2)-(x-2))^2$$

$$\log((x+1)^2 / ((x-2)(x-1))^2$$

$$\log\left(\frac{(x+1)^2}{(x-2)(x-1)}\right)^2=2(\log(x+1)^2-\log(x-2)(x-1))$$

$$=2(2\log(x+1)-\log(x-2)-\log(x-1))=$$ $$=4\log(x+1)-2\log(x-2)-2\log(x-1)$$

-