# Is there a variable that makes 2log(x-3y)=log2+logy?

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You want to solve this equation for x.

2log(x-3y) = log2+logy

Now log2+logy = log(2y) and 2log(x-3y) = log((x-3y)^2) so to have this equality we must have

(x-3y)^2 = 2y Expanding the left side we get

x^2-6xy+9y^2 = 2y Putting this in standard form we get

x^2 - 6xy + 9y^2 - 2y = 0

We could use the quadradic formula to solve for x

`x = (-(-6y)+-sqrt(36y^2-4(1)(9y^2-2y)))/2 = (6y+-sqrt(8y))/2 = 3y+-sqrt(2y)`

Note that `sqrt(8y) = sqrt(4*2y) = sqrt(4)sqrt(2y) = 2sqrt(2y)`

So there are two answers, but `3y-sqrt(2y)-3y = -sqrt(2y)` and that is not allowed in a log because you do not get a real number so the only solution is

`x=3y+sqrt(2y)`