Given that x,y>0 find x/y if x+6y>0 2 (lgx - lgy) = lg (x+ 6y) -lg y

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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

 

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