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Math Help - Simultaneous ln and e equation!

  1. #1
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    Simultaneous ln and e equation!

    e^2y - x + 2 = 0
    ln(x+3) - 2y - 1 = 0

    need help on how to start solving this! thanks
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, jaderberg!

    This is not pretty . . .


    . e^{2y} - x + 2 .= .0
    ln(x + 3) - 2y - 1 .= .0

    We have: .e^{2y} .= .x - 2 - - [1]
    . . . . . . .ln(x + 3) .= .2y + 1 . [2]

    Equation [1] becomes: -ln(x - 2) .= .2y
    . .Subtract it from [2]: .ln(x + 3) .= .2y + 1

    We have: .ln(x + 3) - ln(x - 2) .= .1 . . ln(x + 3) .= .1 + ln(x - 2)

    . . ln(x + 3) .= .ln(e) + ln(x - 2) . . ln(x + 3) .= .ln[e(x - 2)]

    Un-log both sides: .x + 3 .= .ex - 2e . . ex - x .= .2e + 3

    Factor: .x(e - 1) .= .2e + 3

    . . Hence: . x .= .(2e + 3)/(e - 1)


    Substitute into [1]: .e^{2y} .= .(2e + 3)/(e - 1) - 2 .= .5/(e - 1)

    . . Hence: .2y .= .ln[5/(e - 1)] . . y .= .½·ln[5/(e - 1)]

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