Given that x> 0 and y >0.

Also, given that x+ 6y > 0

We need to find x/y.

2 (lgx - lgy) = lg (x+ 6y) -lg y.

We know from logarithm properties that log(a) - log(b) = log (a/b)

==> 2*log (x/y) = log (x+6y)/y

==> Also, we know that a*log b = log b^a

==> log (x/y)^2 = log (x+6y)/y

Now that the logs are equal then the bases are equal.

==> (x/y)^2 = (x+6y)/y

==> (x/y)^2 = (x/y) + 6)

==> (x/y)^2 - (x/y) - 6 = 0

Let us assume that x/y = u

==> u^2 - u - 6 = 0

We will factor.

==> (u-3)(u+2) = 0

==> u1 = 3 ==> x/y = 3

==> u2 = -2 ==> x/y = -2 ( Not valid because both x and y are positive.

**Then the only answer is :**

**==> x/y = 3**

We don't have to impose constraints of existence, since x + 6y>0 and x,y>0 (from enunciation).

We'll solve the given identity.

We'll apply the quotient rule of logarithms both sides:

2 lg (x/y) = lg [(x+6y)/y]

We'll apply power rule to the left side:

lg (x/y)^2 = lg [(x+6y)/y]

Since the bases are matching, we'll have:

(x/y)^2 = [(x+6y)/y]

(x/y)^2 = x/y + 6

We'll note x/y = t

t^2 = t + 6

We'll substract t + 6 both sides:

t^2 - t - 6 = 0

We'll apply quadratic formula:

t1 = [1 + sqrt(1 + 24)]/2

t1 = (1 + 5)/2

t1 = 3

t2 = -2

x/y = 3

and x/y = -2, but x and y are positive so it is impossible that their quotient to be negative.

**The answer is: x/y = 3.**