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Math Help - Initial Population

  1. #1
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    Initial Population

    The question goes: The population of bacteria grows at a rate proportional to the number of bacteria at time t. After 3 hours it is observed that 400 bacteria are present. After 10 hours 2000 bacteria are present. What is the initial number of bacteria?

    y=ce^kt
    y(0)=?
    y(3)=400
    y(10)=2000

    I don't understand how I can solve for the two unknowns: k and c
    ex. 400=ce^k(3)
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  2. #2
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    Quote Originally Posted by latavee View Post
    The question goes: The population of bacteria grows at a rate proportional to the number of bacteria at time t. After 3 hours it is observed that 400 bacteria are present. After 10 hours 2000 bacteria are present. What is the initial number of bacteria?

    y=ce^kt
    y(0)=?
    y(3)=400
    y(10)=2000

    I don't understand how I can solve for the two unknowns: k and c
    ex. 400=ce^k(3)
    and 2000=ce^{k(10)} \Rightarrow c = \frac{2000}{e^{k(10)} }

    with your work 400=ce^{k(3)}\Rightarrow c = \frac{400}{e^{k(3)} }

    gives us \frac{2000}{e^{k(10)} } = \frac{400}{e^{k(3)} }

    \frac{2000}{400} = \frac{e^{k(10)} }{e^{k(3)} }

    5 = e^{k(10)-k(3)}

    5 = e^{10k-3k}

    5 = e^{7k}


    You can now solve for k, which will then give you c.
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  3. #3
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    Thanks!!

    Wow! I never thought about it that way, thanks! So when I plug my k back to find my c, I will actually have to find c1 and c2 correct?
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    Hi all, I am a new member of forum
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  5. #5
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    Quote Originally Posted by latavee View Post
    Wow! I never thought about it that way, thanks! So when I plug my k back to find my c, I will actually have to find c1 and c2 correct?
    Nope, just c, I don't see any others

    After you have k and c you then sub t=0 into the equation for initial population.
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