If #x^2# + #y^2# = #3xy#, 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)#

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

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

Eliminate #log# on both sides
#(x-y) = (xy)^(1/2)#

Square both sides
#(x-y)^2 = ((xy)^(1/2))^2#
#(x-y)(x-y) = xy#
#x^2 - 2xy + y^2 = xy#

Add #2xy# to both sides
#x^2 + y^2 = 3xy#