precalculus

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solve the logarithmic equation . express solution in exact form

log5(x-9)+log5(x+4)=1+log5(x-5)

  • precalculus -

    1 = log5(5), so we have

    log5(x-9)+log5(x+4)=log5(5)+log5(x-5)

    log5[(x-9)(x+4)] = log5[5(x-5)]

    raise 5 to the powers, and we have

    (x-9)(x+4) = 5(x-5)
    x^2 - 5x - 36 = 5x - 25
    x^2 - 10x - 11 = 0
    (x-11)(x+1) = 0

    Solutions are 11,-1

    However, -1 does not fit the original equation: log of negatives are undefined.

  • precalculus -

    log5(x-9)+log5(x+4)=1+log5(x-5)
    log5(x-9)+log5(x+4)=log5(5)+log5(x-5)
    log5[(x-9)(x+4)] = log5[5(x-5)}
    (x-9)(x+4) = 5(x-5)
    x^2 - 5x - 36 = 5x - 25
    x^2 - 10x - 9 = 0
    x = (10 ± √136)/2 = appr. 10.83 or -.83
    but for each of the above to defined, x > 9

    so x = (10 + √136)/2 = 5 + √34

    check my arithmetic

  • Steve had right equation, precalculus -

    I have an error in my equation...

    x^2 - 10x - 9 = 0 should be
    x^2 - 10x - 11 - 0 , just like Steve had

    then (x-11)(x+1) = 0
    x = 11 or x = -1

    so x = 11

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