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Math Help - fun with natural logs....

  1. #1
    Senior Member
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    fun with natural logs....

    ok so we have the problem...

    ln N - ln(N-2000)= t + C

    so i tried to say that ln a = ln B = ln (a/B)

    so i got ln(N/N-2000)= t + c

    so then i raised both sides to e and got

    N/N-2000 = e^(t+c)

    my final answer was -2000e^(t+c)/(1-e^(t+c))

    my book got the same thing except their answer was positive, i dont see how this could be possible? can someone tell me where i went wrong
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  2. #2
    Super Member Quacky's Avatar
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    Quote Originally Posted by slapmaxwell1 View Post
    ok so we have the problem...

    ln N - ln(N-2000)= t + C

    so i tried to say that ln a = ln B = ln (a/B)

    so i got ln(N/N-2000)= t + C

    so then i raised both sides to e and got

    N/N-2000 = e^(t+C)

    my final answer was -2000e^(t+c)/(1-e^(t+c))

    my book got the same thing except their answer was positive, i dont see how this could be possible? can someone tell me where i went wrong
    You haven't done anything wrong at this stage; the error lies elsewhere with your working or with the textbook. What is the original question?

    N=e^{t+C}(N-2000)

    N=Ne^{t+C}-2000e^{t+C}

    N-Ne^{t+C}=-2000e^{t+C}

    N(1-e^{t+C})=-2000e^{t+C}

    N=\dfrac{-2000e^{t+C}}{1-e^{t+C}}
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  3. #3
    MHF Contributor

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    Notice that e^{t+ C}= ce^y where c= e^C.

    Also, you can multiply both numerator and denominator by -1 to get
    \frac{2000ce^t}{ce^t- 1}

    Perhaps that is the "positive" answer you refer to.
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