If x^2x2 + y^2y2 = 3xy3xy, show that #log (x-y) = 1/2(log x + log y) How do you solve this?

1 Answer
May 6, 2018

Solve by Condensing the Right Side and using the Properties of Logs.

Explanation:

log (x-y) = 1/2(log x + log y)log(xy)=12(logx+logy)

Condense Right Side
log (x-y) = 1/2log(xy)log(xy)=12log(xy)

Coefficient becomes the Power of (xy)(xy)
log (x-y) = log(xy)^(1/2)log(xy)=log(xy)12

Eliminate loglog on both sides
(x-y) = (xy)^(1/2)(xy)=(xy)12

Square both sides
(x-y)^2 = ((xy)^(1/2))^2(xy)2=((xy)12)2
(x-y)(x-y) = xy(xy)(xy)=xy
x^2 - 2xy + y^2 = xyx22xy+y2=xy

Add 2xy2xy to both sides
x^2 + y^2 = 3xyx2+y2=3xy