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Exponential growth problem

  1. Dec 28, 2015 #1
    1. The problem statement, all variables and given/known data
    There are initially two cells seen through a microscope. every hour, a computer was tasked with taking a cell count through the microscope. Taking the sum of all the cell counts after the 7th hour, the total cell count was 280 cells. Find the exponential growth rate.

    2. Relevant equations
    $$P = P_0\sum_{t = 1}^{7}{e^{kt}}$$
    $$P_0 = 2$$
    3. The attempt at a solution

    $$2\sum_{t=1}^{7}e^{kt} = P$$
    $$\sum_{t=1}^{7}e^{kt} = 280/2$$
    $$ln(140) = \sum_{t=1}^{7}(kt) = k(\sum_{t=1}^{7}t) = k*(28)$$
    $$k = \frac{ln(140)}{28}$$
    $$ k = 0.176487$$

    if you plug this into the original equation:

    $$P_0\sum_{t = i}^{7}{e^{0.176487t}}$$
    this sum from 1 to 7 does not equal my final P value, what am i missing??
     
  2. jcsd
  3. Dec 28, 2015 #2

    haruspex

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    The log of a sum is not the sum of the logs. You need to sum the series first.
     
  4. Dec 28, 2015 #3

    SammyS

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    The first "Relevant Equation" ##\displaystyle \ P = P_0\sum_{t = 1}^{7}{e^{kt}} \ ## is incorrect for exponential growth.

    It's ##\displaystyle \ P(t) = P_0\ e^{kt} \ ##
     
  5. Dec 28, 2015 #4

    haruspex

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    I believe iScience is using P for the cumulative total of observations, not the population at a time step.
     
  6. Dec 29, 2015 #5

    Ray Vickson

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    As has already been pointed out: you cannot distribute the "log" over the "sum".
    Your basic problem is to solve the equation
    [tex] 2(r + r^2 + \cdots + r^7) = 280, \; (r = e^k) [/tex]
    You can solve for ##r## first, then get ##k## from that. However, solving for ##r## involves solving a 7th degree polynomial (and even doing the 7-term sum first does not help), so you must fall back on graphical/numerical solution methods. There will not be any nice formulas you can apply.
     
  7. Dec 30, 2015 #6
    This is the sum of a geometric progression. That might simplify things a little.
     
  8. Dec 30, 2015 #7

    Ray Vickson

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    As I said in # 5, even doing the sum does not help much; it replaces a 7th degree polynomlal equation by an 8th degree one!
     
  9. Dec 30, 2015 #8
    But, the 8th degree equation only involves r8 and r. I solved the equation $$\frac{r^8-1}{r-1}=140$$ in 4 iterations. My first guess was r = 1.75. So, my sequence of guesses was
    1.75
    1.8
    1.81
    1.805

    This all took a total of less than 5 minutes on my calculator.

    Chet
     
  10. Dec 30, 2015 #9

    Ray Vickson

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    Well, of course there are numerous ways of solving that problem (or the non-summed one) numerically, and some are faster and/or easier than others. The main point I wanted to get across to the OP was that some such method must be used, and that there will be no nice "closed-form" formula that can be used to get the solution. However, it looks like he/she may have abandoned the thread.
     
  11. Dec 30, 2015 #10
    Thanks Ray. Anyway, it's a good thing he left because I summed the geometric progression incorrectly anyway. I should have solved:
    $$\frac{r(r^7-1)}{r-1}=140$$

    Oh well.
     
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