if x^2+y^2=3xy then prove that 2log(x-y)=logx-logy
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log (x+y/ 5) = 1/2 (logx + logy)
log (x+y/ 5) = 1/2 (log xy)
log (x+y/ 5) = log (xy)^(1/2)
(x+y/ 5) = (xy)^(1/2)
(x+y/ 5)^2 = xy
(x² + y² + 2xy) / 25 = xy
x^2 + y^2 = 23xy.
log (x+y/ 5) = 1/2 (log xy)
log (x+y/ 5) = log (xy)^(1/2)
(x+y/ 5) = (xy)^(1/2)
(x+y/ 5)^2 = xy
(x² + y² + 2xy) / 25 = xy
x^2 + y^2 = 23xy.
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