0
$\begingroup$

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 $\big((x-1)(x-2)\big)^2$?

  • 3
    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 $4\log(x+1)-2\log((x-1)(x-2))$).2012-11-18

2 Answers 2