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

JP
Answered by Joe P. Maths tutor

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