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Math Help - Insect population growth problem

  1. #1
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    Insect population growth problem

    An insect population is increasing exponentially. 4 days after the population was established there are 400 insects, and after 3 more days, there are 1600 insects. How many insects will there be 10 days after the population was established? Thank you in advance.
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    Quote Originally Posted by lucky13 View Post
    An insect population is increasing exponentially. 4 days after the population was established there are 400 insects, and after 3 more days, there are 1600 insects. How many insects will there be 10 days after the population was established? Thank you in advance.
    Use the model P(t) = P_0e^{rt} and find P_0 and r via substitution of (4,400) and (7,1600)

    After this find P(10) and you will have the answer.
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    Im not sure I understand, Where do I substitute those numbers to?
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    Quote Originally Posted by lucky13 View Post
    Im not sure I understand, Where do I substitute those numbers to?
    In this case (4,400) = (t,P(t))

    We get 400 = P_0e^{4r}

    Do the same for the other point.
    What do you get?
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  5. #5
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    1600=Poe^7r?
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    Quote Originally Posted by lucky13 View Post
    1600=Poe^7r?
    Yep, now solve your equation simultaneously with

    400 = P_0e^{4r}
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    Solve it simultaneusly?
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  8. #8
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    Quote Originally Posted by lucky13 View Post
    Solve it simultaneusly?
    You have...

    P_0 e^{7r} = 1600
    P_0 e^{4r} = 400

    Divide the top equation by the bottom one.

    \frac{P_0 e^{7r}}{P_0 e^{4r}} = \frac{1600}{400}

    => e^{(7-4)r} = 4

    => r = \frac{2}{3}\ln(2)

    Substitute that into P_0 e^{4r} = 400 to get...

    P_0 \cdot 4 \cdot 2^{2/3} = 400

    => P_0 = 50 \cdot 2^{1/3}

    You can simplify stuff if ya like, I've left it all in a kind of messy form.

    Then solve...

    P_{10} = P_0 e^{10r}
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    Quote Originally Posted by Deadstar View Post
    You have...

    P_0 e^{7r} = 1600
    P_0 e^{4r} = 400

    Divide the top equation by the bottom one.

    \frac{P_0 e^{7r}}{P_0 e^{4r}} = \frac{1600}{400}

    => e^{(7-4)r} = 4

    => r = \frac{2}{3}\ln(2)

    Substitute that into P_0 e^{4r} = 400 to get...

    P_0 \cdot 4 \cdot 2^{2/3} = 400

    => P_0 = 50 \cdot 2^{1/3}

    You can simplify stuff if ya like, I've left it all in a kind of messy form.

    Then solve...

    P_{10} = P_0 e^{10r}
    P_{10} =50 \cdot 2^{1/3}e^{10(\frac{2}{3}\ln(2))}

    =50 \cdot 2^{1/3}(e^{ln(2)})^{\frac{20}{3}}

    =50 \cdot 2^{1/3}(2)^{\frac{20}{3}}

    =50 \cdot 2^7

    =6400
    Last edited by ione; April 23rd 2010 at 10:27 PM.
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