Question

if log[×+y/3] =1/2(logx+logy) then find the value of x/y-y+x

Answer 1

log{{x+y}/3}=1/2{logx+log Y}
log(x+y) - log(3) = 1/2(log xy)
log(xy) - log(2) = 1/2(log xy)
2xy - 4 = xy
xy = 4
now

x/y + y/x =
(x^2 + y^2) / xy =
(x^2 + y^2) / 4
we have two cases
(4, 1) and (2,2) are factors of 4
1) (16 + 1) / 4 = 17/4
2) (4 + 4) / 4 = 2