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Math Help - Bacteria Problem

  1. #1
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    Bacteria Problem

    This poblem Involves bacteria...
    there are initially 4000 bacteria per mililiter in a sample. after one hour their concentration increases to 6000 bacteria per mililiter.
    a) what is the general exponetial equasion?
    b) How many bacteria will there be after 2.5 hours
    c) how many hours does it take to reach 8000 bacteria per mililiter.



    so far i have isolated that the equasion N(t)=N e^kt is used to solve it
    with N(t) being the final number
    N being the initial number (4000)
    k being a positive constant
    and t being the number of hours

    from here I have 6000= 4000 e^k*1

    any help would be greatly apreciated thank you
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mazdatuner
    This poblem Involves bacteria...
    there are initially 4000 bacteria per mililiter in a sample. after one hour their concentration increases to 6000 bacteria per mililiter.
    a) what is the general exponetial equasion?
    b) How many bacteria will there be after 2.5 hours
    c) how many hours does it take to reach 8000 bacteria per mililiter.



    so far i have isolated that the equasion N(t)=N e^kt is used to solve it
    with N(t) being the final number
    N being the initial number (4000)
    k being a positive constant
    and t being the number of hours

    from here I have 6000= 4000 e^k*1

    any help would be greatly apreciated thank you
    a) To get k:
    6000=4000e^k

    6000/4000=3/2=e^k

    ln(3/2)=ln(e^k)=k.
    I don't have my calculator on me (and I'm too lazy to dig out the Window's version.) You don't HAVE to have a decimal form for this, but that generally seems to be the standard way to express it.

    b) N(2.5)=4000e^{k*2.5}. Just calculate it using the k value from a)

    c) N(t) = 8000
    8000=4000e^{kt}

    2 = e^{kt}

    ln2 = ln(e^{kt})=kt

    t = ln2/k.

    -Dan
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