Find values of y such that: log2(11y–3)–log2(3) –2log2(y) = 1

NB.: Treat all log as log2 for purpose of formatting log(x) - log(z) = log(x/z) alog(b) = log(b^a) log((11y - 3)/3) - log(y^2) = 1 log((11y - 3)/3y^2) = 1 11y - 3 / 3y^2 = 2^1 11y - 3 = 6y^2 6y^2 - 11y + 3 = 0 (3y - 1) (2y - 3) = 0 y = 1/3 or 1.5

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Answered by Shrinivas A. Maths tutor

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