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Math Help - logarithm tomfooleries, please help

  1. #1
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    logarithm tomfooleries, please help

    hi, me again, how to solve this, logx + lnx = 10, does anyone know? thank you for your time and help. EDIT: also how do logarithms help to solve simultaneous eqautions, such as x+y=5 and xy=6, or xy=24 and (e^x)(e^y)=e^10 etc. thanks anyone
    Last edited by captainlewis; May 3rd 2007 at 09:18 AM.
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  2. #2
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    Hello, captainlewis!

    Solve: .log(x) + ln(x) .= .10
    Use the Base-Change formula . . .

    . . .ln(x)
    . . ------- + ln(x) .= .10
    . . ln(10)

    Multiply by ln(10): .ln(x) + ln(10)Ěln(x) .= .10Ěln(10)

    Factor: . ln(x)Ě[1 + ln(10)] .= .10Ěln(10)

    . . . . . . . . . . . . .10Ěln(10)
    Then: . ln(x) .= .------------ . .6.97
    . . . . . . . . . . . . 1 + ln(10)


    Therefore: .x .= .e^6.97 . .1064.2




    Solve: .xy .= .24, .(e^x)(e^y) .= .e^10
    The second equation is: .e^{x+y} .= .e^10 . . x + y .= .10 .[1]

    The first equation is: .xy = 24 . . y = 24/x .[2]


    Substitute [2] into [1]: .x + 24/x .= .10

    . . which gives us a quadratic: .x▓ - 10x + 24 .= .0

    . . which factors: .(x - 4)(x - 6) .= .0

    . . and has roots: .x .= .4, 6

    Substitute into [2]: .y .= .6, 4


    Therefore: .(x,y) .= .(4, 6) and (6, 4)

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  3. #3
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    thank you so much

    thank you so much, this has solved my problems.
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