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#.