Question #89e4a
1 Answer
Apr 9, 2017
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)))#
