Given that x²+y=5xy show that

Question

A) 2log(x+y/√7)=logx+logy

B)log(x-y/√3)=½(logx+logy)

Answer 1

Fixing your typo, we have
x^2 + y^2 = 5xy
equating A and B,
log (x+y)^2/7 = log(x-y)^2/3
3(x+y)^2 = 7(x-y)^2
3x^2 + 6xy + 3y^2 = 7x^2 - 14xy + 7y^2
4x^2 - 20xy + 4y^2 = 0
x^2 + y^2 = 5xy

So see what you can do with A and B separately.