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Math Help - Exponential Growth

  1. #1
    Member CalcGeek31's Avatar
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    Exponential Growth

    Here is the question:
    Under ideal conditions, a certain bacteria population is known to double every three hours. Initially there are 100 bacteria.

    I need to figure out how many bacteria there are at 15 hours and at t hours... I really need help with this please help.
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  2. #2
    Eater of Worlds
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    We have 200 bacteria in 3 hours because it doubles.

    200=100e^{3k}

    Solve for k=\frac{ln(2)}{3}

    A=100e^{t\frac{ln(2)}{3}}

    Let t=15.
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  3. #3
    Member CalcGeek31's Avatar
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    Okay thank you but now I have to chang it completely to ln/log format to get something = to t right?
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  4. #4
    Eater of Worlds
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    You can solve for t if your wish, but the equation I gave should suffice for the number of bacteria given any t.
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  5. #5
    Member CalcGeek31's Avatar
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    unfortunately I have to solve for t according to my BC Calc Packet
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  6. #6
    Eater of Worlds
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    You can also express e^{\frac{ln(2)t}{3}}=2^{\frac{t}{3}}

    A=100e^{\frac{ln(2)t}{3}}

    t=\frac{3ln(\frac{A}{100})}{ln(2)}
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