What is the ratio x/y whether 2log(2x-3y)=logx+logy ?

We'll apply the power rule of logarithms for the term from
the left side:

2*log (2x-3y) = log
(2x-3y)^2

Since the bases of logarithms from the sum from
the right side are matching, we'll apply the product
rule:

logx+logy = log
(x*y)

We'll re-write the
identity:

log (2x-3y)^2 = log
(x*y)

Since the bases are matching, we'll apply one to one
rule:

(2x-3y)^2 = x*y

We'll
expand the square form the left:

4x^2 - 12x*y + 9y^2 - x*y
= 0

4x^2 - 13x*y + 9y^2 =
0

We'll divide by
y^2:

4(x/y)^2 - 13(x/y) + 9 =
0

We'll replace the fraction x/y by
t:

4t^2 - 13t + 9 = 0

We'll
apply the quadratic formula:

t1 = [13+sqrt(169 -
144)]/8

t1 = (13+5)/8

t1 =
18/8

t1 = 9/4

t2 =
(13-5)/8

t2 =
1

The possible values of the ratio x/y are:
x/y = 9/4 and x/y = 1.