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Math Help - [SOLVED] exponential function

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    Post [SOLVED] exponential function

    I've been trying to do this exercise for a while , and the only result I've gotten is 440. Can someone help me. Thanks in advance!

    At the beginning of an experiment, a culture contains 200 bacteria. An hour later there are 205 bacteria. Assuming that the bacteria grow exponentially, How many will there be after 2 day?
    Last edited by vance; February 20th 2009 at 08:30 PM.
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    Quote Originally Posted by vance View Post
    I've been trying to do this exercise for a while , and the only result I've gotten is 440. Can someone help me. Thanks in advance!

    At the beginning of an experiment, a culture contains 200 bacteria. An hour later there are 205 bacteria. Assuming that the bacteria grow exponentially, How many will there be after 2 day?
    The quantity of bacteria at time t should be of the form f(t)=ke^{at}. Let's use hours as the unit for t\text.

    From the problem statement, we have

    f(0) = 200\Rightarrow ke^{0a}=200\Rightarrow k=200\text.

    So f(t) = 200e^{at}\text. Then, from the second condition given, we know that

    f(1)=205\Rightarrow200e^a=205\Rightarrow e^a=\frac{205}{200}=\frac{41}{40}

    \Rightarrow a=\ln\left(\frac{41}{40}\right)\text.

    Thus, we have f(t)=200e^{t\ln(41/40)}=200\left(\frac{41}{40}\right)^t, so

    f(48) = 200\left(\frac{41}{40}\right)^{48}\approx654.30\te  xt{ bacteria.}
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    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by vance View Post
    I've been trying to do this exercise for a while , and the only result I've gotten is 440. Can someone help me. Thanks in advance!

    At the beginning of an experiment, a culture contains 200 bacteria. An hour later there are 205 bacteria. Assuming that the bacteria grow exponentially, How many will there be after 2 day?
    First, remember that your times must be in the same units. In other words, either use days or hours. You are given times of 1 hour and 2 days. You can either calculate how many days 1 hour is equal to or how many hours are in two days. I would personally use the latter.

    f(t) = Pe^{rt}

    f(0) = 200e^{0r} = 200

    You need to figure out the rate.

    205 = 200e^{1r}

    1.025 = e^{r}

    r = ln(\frac{205}{200})

    So for 48 hours:

    f(48) = 200e^{(\frac{205}{200})(48)}

    f(48) = 654.3
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    Thanks Guys!!!
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    You don't HAVE to use "e" for exponential problems. Saying that the growth is exponential tells us that f= ab^t. Since b^0= 1 for all b, f(0)= a= 200. Taking t in hours, when t= 1 we have [tex]f(t)= 200b= 205[tex] so b= \frac{205}{200}= \frac{41}{40}. After 48 hours, there are
    f(48)= 200b^{48}= 200b^{48}= 200(\frac{41}{40})^{48}= 200(3.27148956)= 654 bacteria.
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