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Math Help - need help growth problem

  1. #1
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    need help growth problem

    given y=Ce^kt where C is the initial value

    The count in a bacteria culture was 5000 after 15 min and 40000 after 1 hour

    What is the initial size of the culture?


    Thanx in advanced
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by justin016 View Post
    given y=Ce^kt where C is the initial value

    The count in a bacteria culture was 5000 after 15 min and 40000 after 1 hour

    What is the initial size of the culture?


    Thanx in advanced
    Use the two conditions to solve for C and k

    Is time measured in hours or minutes?

    5000 = C e^(k15)
    40000 = C e^(k60)

    Divide the second by the first, side by side:
    8 = e^(60k-15k)

    8 = e^(45k)
    ln 8 = 45k

    k =ln8 /45

    Plug into one of the conditions above to find C. C is the original population, because it is the y-value at t=0.
    Good luck!!
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  3. #3
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    i plugged k= ln8/45 into Ce^k15=5000 to find C ,and i get 12.2877

    The answer i am looking for is 2500. What did i do wrong?
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  4. #4
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by justin016 View Post
    i plugged k= ln8/45 into Ce^k15=5000 to find C ,and i get 12.2877

    The answer i am looking for is 2500. What did i do wrong?
    Try units in hours. I assumed minutes in the notes above.
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  5. #5
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    Quote Originally Posted by apcalculus View Post
    Try units in hours. I assumed minutes in the notes above.
    I still get the same answer
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  6. #6
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    Quote Originally Posted by justin016 View Post
    i plugged k= ln8/45 into Ce^k15=5000 to find C ,and i get 12.2877

    The answer i am looking for is 2500. What did i do wrong?
    If you don't post your working, how can you expect anyone to answer this? And how can you expect to learn from your mistake(s).

    Quote Originally Posted by justin016 View Post
    I still get the same answer
    A start:

    5000 = C e^{\left( \frac{\ln 8}{45}\right) 15} = C e^{\frac{\ln 8}{3}} = C e^{\ln 2}

    where you get that last expression by applying a well known log rule.
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  7. #7
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    Ce^ln8/3=5000
    Ce^ln2=5000
    C2=5000
    C=5000/2
    C=2500

    I got confuse where you changed from ln8/3=ln2. How did you do that?
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  8. #8
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    Quote Originally Posted by justin016 View Post
    Ce^ln8/3=5000
    Ce^ln2=5000
    C2=5000
    C=5000/2
    C=2500

    I got confuse where you changed from ln8/3=ln2. How did you do that?
    \frac{\ln 8}{3} = \frac{1}{3} \ln 8 = \ln 8^{1/3} using the usual log rule a \ln b = \ln b^a.

    And you should know that 8^{1/3} = \sqrt[3] 8 = 2.
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    \frac{\ln 8}{3} = \frac{1}{3} \ln 8 = \ln 8^{1/3} using the usual log rule a \ln b = \ln b^a.

    And you should know that 8^{1/3} = \sqrt[3] 8 = 2.

    Okay i got it, thanx alot
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