# Question 89e4a

Apr 9, 2017

$\log \left(\frac{\left(x + 3\right) \left(x - 1\right)}{\left(x - 2\right) \left(x + 6\right) \left(x + 1\right)}\right)$

#### Explanation:

Using the $\textcolor{b l u e}{\text{law of logarithms}}$

• logx-logy=log(x/y)#

$\text{here } x = \frac{{x}^{2} + 2 x - 3}{{x}^{2} - 4} = \frac{\left(x + 3\right) \left(x - 1\right)}{\left(x - 2\right) \left(x + 2\right)}$

$\text{and } y = \frac{{x}^{2} + 7 x + 6}{x + 2} = \frac{\left(x + 6\right) \left(x + 1\right)}{x + 2}$

$\Rightarrow \frac{x}{y} = \frac{\left(x + 3\right) \left(x - 1\right)}{\left(x - 2\right) \cancel{\left(x + 2\right)}} \times \frac{\cancel{\left(x + 2\right)}}{\left(x + 6\right) \left(x + 1\right)}$

$\textcolor{w h i t e}{\Rightarrow \frac{x}{y}} = \frac{\left(x + 3\right) \left(x - 1\right)}{\left(x - 2\right) \left(x + 6\right) \left(x + 1\right)}$

$\Rightarrow \log \left(\frac{{x}^{2} + 2 x - 3}{{x}^{2} - 4}\right) - \log \left(\frac{{x}^{2} + 7 x + 6}{x + 2}\right)$

$= \log \left(\frac{\left(x + 3\right) \left(x - 1\right)}{\left(x - 2\right) \left(x + 6\right) \left(x + 1\right)}\right)$