Find the values of y such that log2(11y-3)-log2(3)-2log2(​y) = 1

Find the values of y such that log₂(11y-3)-log₂3-2log₂y = 1

Power law: 2log₂y = log₂y²

Product law: log₂(11y-3) - log₂3 - log₂y2 =  log₂(11y-3) - log₂(3y²)

Quotient law:  log₂(11y-3) - log₂(3y²) =  log₂(11y-3/3y²)

log₂(11y-3/3y²) = 1

So, 11y-3/3y² = 2¹ = 2

11y - 3 = 2(3y²) = 6y²

0 = 6y²-11y+3

0 = (3y-1)(2y-3)

y = 1/3 or y = 3/2