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Solve simultaneously log (x-1)+2logy=2log3 and logx+logy=log6

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  1. From the first, using log rules:
    log (x-1)+2logy=2log3
    log (x-1)+log(y^2)=log(3^2)
    log( (x-1)(y^2) ) = log 9
    (x-1)y^2 = 9

    from the 2nd:
    logx+logy=log6
    log(xy) = log6
    xy = 6 or x = 6/y

    now use substitution,
    (6/y - 1)(y^2) = 9
    6y - y^2 - 9 = 0
    y^2 - 6y + 9 = 0
    (y - 3)^2 = 0
    y - 3 = 0
    y = 3
    then x = 6/3 = 2

    Check with your calculator.

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    Reiny

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