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 $\big((x-1)(x-2)\big)^2$?
Helping turning this into sum/difference logarithm?
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3Since 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 $4\log(x+1)-2\log((x-1)(x-2))$). – 2012-11-18
2 Answers
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.
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0@KelseyAbreu I have taken into account the entire power $2$ when I wrote "Using the first property, we get that $\log \left( \left( \dfrac{x^2+2x+1}{x^2 - 3x+2} \right)^2\right) = 2 \log \left( \dfrac{x^2+2x+1}{x^2 - 3x+2}\right)"$ – 2012-11-18
$\log((x^2 +2x+1) / (x^2 -3x +2))^2$ $\log((x+1)^2 / (x^2 -2x-(x-2))^2$
$\log((x+1)^2 / (x(x-2)-(x-2))^2$
$\log((x+1)^2 / ((x-2)(x-1))^2$
$\log\left(\frac{(x+1)^2}{(x-2)(x-1)}\right)^2=2(\log(x+1)^2-\log(x-2)(x-1))$
$=2(2\log(x+1)-\log(x-2)-\log(x-1))=$ $=4\log(x+1)-2\log(x-2)-2\log(x-1)$