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Math Help - Exponential function word problems?

  1. #1
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    Question Exponential function word problems?

    I really don't enjoy word problems. Can you give me a hand with the following? Once it's in an equation I'm pretty good..

    5. For a biology experiment, the number of cells present is 1000. After 4 hours, the count is estimated to be 256000. What is the "doubling period" of the cells?
    * How do you derive an equation from this? I tried and got x= 1/2 as my answer. Am i anywhere close?

    6. A certain bacteria strain divides every 0.5 hours. If 500 bacteria are present in a culture, how many will be there
    a) After 3 hours?
    B) After 6 hours?
    C) After n hours?

    10. A certain strain of yeast cell doubles every 20 minutes. If there were 350 initially, how many will there be in 3 hours?

    Again guys, I could really use the help.
    Thanks so much!
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  2. #2
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    Quote Originally Posted by brittt-- View Post
    I really don't enjoy word problems. Can you give me a hand with the following? Once it's in an equation I'm pretty good..

    5. For a biology experiment, the number of cells present is 1000. After 4 hours, the count is estimated to be 256000. What is the "doubling period" of the cells?
    * How do you derive an equation from this? I tried and got x= 1/2 as my answer. Am i anywhere close?

    6. A certain bacteria strain divides every 0.5 hours. If 500 bacteria are present in a culture, how many will be there
    a) After 3 hours?
    B) After 6 hours?
    C) After n hours?

    10. A certain strain of yeast cell doubles every 20 minutes. If there were 350 initially, how many will there be in 3 hours?

    Again guys, I could really use the help.
    Thanks so much!
    As it's exponential growth your equation will be of the form N = Ae^{kt} where N is the number, A is a constant (which I will soon denote N_0), k is the growth constant (dimensions: 1/time) and t is time.

    To find A we use the initial condition; that is how many are at t=0. It becomes clear that N=A and so we let A=N_0 and define N_0 as the number at the start.

    In your case you want to find k so rearranging my main equation gives

    k = \frac{1}{t} \ln{(\frac{N_0}{N})} = \frac{1}{4}ln(\frac{256000}{1000})

    You could number crunch to find k but it can be simplified since 256 = 2^8 and a factor of 1000 cancels to get

    k  = 2ln(2).

    Now we have our constant we can find out a relationship between doubling time and the constant k.

    From the first equation: N = N_0e^{kt} at doubling time N = 2N_0 and so t_{double} = \frac{ln(2)}{k}

    In your case k  = 2ln(2) so t_{double} = \frac{ln(2)}{2ln(2)} = \frac{1}{2} hour which makes your answer almost correct, you just need a unit.

    You can use the main equation N = N_0e^{kt} for all these, just changing the values as necessary
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  3. #3
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    Quote Originally Posted by brittt-- View Post
    I really don't enjoy word problems. Can you give me a hand with the following? Once it's in an equation I'm pretty good..

    5. For a biology experiment, the number of cells present is 1000. After 4 hours, the count is estimated to be 256000. What is the "doubling period" of the cells?
    * How do you derive an equation from this? I tried and got x= 1/2 as my answer. Am i anywhere close?
    A=Pe^{rt}

    A = the ending amount of cells
    P = the beginning amount of cells
    r = rate of growth
    t = time

    256000=1000e^{4r}

    256=e^{4r}

    \ln 256=4r

    \frac{\ln 256}{4}=r

    We won't approximate here in order to avoid rounding errors.


    To find out when the sample doubles,

    2000=1000e^{\frac{\ln 256}{4}t}

    2=e^{\frac{\ln 256}{4}t}

    \ln 2=\frac{\ln 256}{4}t

    4 \ln 2=\ln 256t

    t=\frac{4 \ln2}{\ln 256}

    t=.5
    Last edited by masters; April 6th 2009 at 04:21 PM. Reason: Revised to avoid rounding errors
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