If $x>y>0$ and $2 log(x-y)=log x + log y$ , then what does $(x/y)$ equal?

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Answer 1 · 2016-08-12T18:33:12

$x/y =1/2(3 + sqrt(5))$

From $2 log(x-y)=log x + log y$ we know

$(x-y)^2=xy$ . Now calling $x = lambda y$ and substituting

$(lambda y-y)^2=lambda y^2$ or supposing $y > 0$

$(lambda-1)^2=lambda$

Solving for $lambda$

$lambda^2-3lambda+1 = 0$

$lambda = 1/2(3 pm sqrt(5))$

so remembering that $x > y$ the feasible solution is $x/y =1/2(3 + sqrt(5))$

Answer 2 · 2016-08-12T18:35:39

$x/y=(3+sqrt5)/2$ .

Given that, $2log(x-y)=logx+logy$

$rArr log(x-y)^2=log(xy)$

$rArr (x-y)^2=xy$

$rArr x^2-3xy+y^2=0$

Dividing by $y^2!=0$ , we get, $(x/y)^2-3(x/y)+1=0$

Hence, $x/y={3+-sqrt((-3)^2-4*1*1)}/2=(3+-sqrt5)/2$

But $x>y>0 rArr x/y>1 rArr x/y=(3-sqrt5)/2<1$ is not admissible.

Therefore, $x/y=(3+sqrt5)/2$ .

If x>y>0x>y>0 and 2 log(x-y)=log x + log y2 log(x-y)=log x + log y, then what does (x/y)(x/y) equal?