Question #89e4a

1 Answer
Apr 9, 2017

log(((x+3)(x-1))/((x-2)(x+6)(x+1)))

Explanation:

Using the color(blue)"law of logarithms"

• logx-logy=log(x/y)

"here "x=(x^2+2x-3)/(x^2-4)=((x+3)(x-1))/((x-2)(x+2))

"and " y=(x^2+7x+6)/(x+2)=((x+6)(x+1))/(x+2)

rArrx/y=((x+3)(x-1))/((x-2)cancel((x+2)))xxcancel((x+2))/((x+6)(x+1))

color(white)(rArrx/y)=((x+3)(x-1))/((x-2)(x+6)(x+1))

rArrlog((x^2+2x-3)/(x^2-4))-log((x^2+7x+6)/(x+2))

=log(((x+3)(x-1))/((x-2)(x+6)(x+1)))