Answer
$\log(\dfrac{(x-3)}{(x+6)(x-2)})\text{ , } x>3$
Work Step by Step
$\log(\dfrac{x^2+2x-3}{x^2-4})-\log(\dfrac{x^2+7x+6}{x+2})$
$\log(\dfrac{x^2+2x-3}{x^2-4})-\log(\dfrac{(x^2+7x+6)(x-2)}{(x+2)(x-2)})$
$\log(\dfrac{x^2+2x-3}{x^2-4})-\log(\dfrac{(x^2+7x+6)(x-2)}{x^2-4})$
Now, we can merge both into a single logarithm
$\log(\dfrac{x^2+2x-3}{x^2-4}\div\dfrac{(x^2+7x+6)(x-2)}{x^2-4})$
We can factor to simplify the expression
$\log(\dfrac{x^2+2x-3}{x^2-4}\cdot\dfrac{x^2-4}{(x^2+7x+6)(x-2)})$
$\log(\dfrac{x^2+2x-3}{(x^2+7x+6)(x-2)})$
$\log(\dfrac{(x-3)(x+1)}{(x+6)(x+1)(x-2)})$
$\log(\dfrac{(x-3)}{(x+6)(x-2)})\text{ , } x>3$
