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Math Help - exp growth/doubling period

  1. #1
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    exp growth/doubling period

    The count in a bacteria culture was 200 after 20 minutes and 1600 after 35 minutes.
    1)What was the initial size of the culture?
    2)Find the doubling period in minutes ?
    3)Find the population after 115 minutes ?

    I was able to solve for number 1, and go it to be 12.4999. I need help setting up the equation for question 2.
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  2. #2
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    i think i got the answer for question 2.

    since k = 0.1386

    2 = e^{0.1386t}
    ln2 = 0.1386t
    t = 5
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by viet View Post
    The count in a bacteria culture was 200 after 20 minutes and 1600 after 35 minutes.
    1)What was the initial size of the culture?
    ok, so we know we are going to use the exponential growth formula, it goes like this.

    A(t) = A_0 e^{rt}

    we have that A(20) = 200

    \Rightarrow 200 = A_0 e^{20r}

    we also have A(35) = 1600

    \Rightarrow 1600 = A_0 e^{35r}

    Let's divide these two equations...

    \Rightarrow \frac {A(35)}{A(20)} = \frac {1600}{200} = \frac {A_0 e^{35r}}{A_0 e^{20r}}

    \Rightarrow 8 = e^{15r}

    \Rightarrow \ln 8 = 15r

    \Rightarrow r = \frac { \ln 8}{15}

    So we are working with A(t) = A_0 e^{ \frac { \ln 8}{15} t}

    Let's use one of the above equations to solve for A_0

    A(20) = 200 = A_0 e^{ \frac { \ln8}{15} \cdot 20}

    \Rightarrow A_0 = \frac {200}{e^{ \frac { \ln 8}{3}4}}

    \Rightarrow A_0 = \frac {200}{8^{ \frac {4}{3}}}

    \Rightarrow A_0 = 12.5

    weird number, check my computation

    2)Find the doubling period in minutes ?
    we want to find t such that we have double the bacteria.

    we can simply solve 2 = e^{ \frac { \ln8}{15}t}

    do you see why? you don't have problems solving that do you?

    oh, i think that's what you did, so i guess you know why



    3)Find the population after 115 minutes ?
    Just solve A(t) = 12.5 e^{ \frac { \ln 8}{15} \cdot 115}
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