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Algebra2

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Use properties of logartithms to simplify and solve. Thanks.

ln(x-3)-ln(11x-23)+in(x-1)=0

  • Algebra2 -

    I believe it may be x=2,x=13

  • Algebra2 -

    Did you test your answers?
    Obviously x= 2 cannot be right since it would make
    the second term undefined,
    ln(22-23) = ln(-1), which is undefined


    using the rules of logs, we get
    ln[(x-3)(x-1)/(11x - 23) ] = 0
    (x-3)(x-1)/(11x - 23) = e^0 = 1
    x^2 - 4x + 3 = 11x - 23
    x^2 - 15x+ 26 = 0
    (x-2)(x-13) = 0
    so x=2 or x=13

    check for x=13
    LS = ln10 - ln120 + ln12
    = ln(10(12)/120) = ln 1 = 0 = RS

    x = 13

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