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Solve this simultaneously log(x-2) + log 2 = 2log y and log( x -3y+3) = 0

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  1. log(x-2) + log 2 = 2log y
    log( x -3y+3) = 0
    removing the logs, this becomes
    2(x-2) = y^2
    x-3y+3 = 1
    since x = 3y-2, we now have
    2(3y-2)-4 = y^2
    y^2-6y+8 = 0
    (y-2)(y-4) = 0
    y = 2 or 4
    so x = 4 or 10

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