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

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  1. log ( x -1 ) + 2 log y = 2 log 3

    log ( x -1 ) = 2 log 3 - 2 log y

    log ( x -1 ) = 2 ( log 3 - log y )

    log ( x -1 ) = 2 log ( 3 / y )

    log ( x -1 ) = log [ ( 3 / y )² ]

    x - 1 = ( 3 / y )²

    x - 1 = 9 / y²

    x = 9 / y² + 1


    log x + log y = log 6

    log ( x ∙ y ) = log 6

    x ∙ y = 6

    x = 6 / y

    Use fact:

    x = x

    9 / y² + 1 = 6 / y

    Multiply both sides by y²

    9 + y² = 6 y

    Subtract 6 y to both sides

    9 + y² - 6 y = 6 y - 6 y

    y² - 6 y + 9 = 0

    Factor the left hand side

    ( y - 3 )² = 0

    y - 3 = √0

    y - 3 = 0

    add 3 to both sides

    y - 3 + 3 = 0 + 3

    y = 3


    x = 6 / y

    x = 6 / 3

    x = 2

    Solution x = 2 , y = 3

    Proof:

    log ( x -1 ) + 2 log y = 2 log 3

    log ( 2 -1 ) + 2 log 3 = 2 log 3

    log 1 + 2 log 3 = 2 log 3

    0 + 2 log 3 = 2 log 3

    2 log 3 = 2 log 3


    log x + log y = log 6

    log 2 + log 3 = log 6

    log ( 2 ∙ 3 ) = log 6

    log 6 = log 6

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    posted by Bosnian

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