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Population growth using logarithims

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data

    A culture begins with 100,000 bacteria and grows to 125,000 bacteria after 20 min. What is the doubling period to the nearest minute?

    2. Relevant equations

    Current=Original(rate)^time

    3. The attempt at a solution

    I can get make the first part out. 125000=100000(rate)^2 I have a feeling its wrong but its as far as i can get.
     
  2. jcsd
  3. Jan 25, 2010 #2

    Char. Limit

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    Gold Member

    Well, yeah. The doubling period is time, and it's supposed to be the thing you're solving for.
     
  4. Jan 25, 2010 #3
    So would my rate be 2? if so then would this be right 120000=100000(2)^t
    1.25=2^t
    log1.25/log2=t?

    What confuses me is when it says doubling period, is that time or the rate?
     
  5. Jan 25, 2010 #4

    Char. Limit

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    Gold Member

    Well, start with this:

    The original is 100,000 and it changes to 125,000 in 20 minutes. So, first solve for the rate, then put in the rate to the equation 200,000=100,000(rate)^(time)
     
  6. Jan 25, 2010 #5
    Well i can get this far but i cant get farther sorry.
    125000=100000(rate)^20
    125000/100000=r^20
    log (125000/100000)=20log r

    I don't know how to go farther, if r had a value and i was solving for time i would have no problem with this.
     
  7. Jan 25, 2010 #6
    alrighhhht worked it out on my own :)

    I figured, why log to find the rate, 20th root it.

    125000/100000=r^20
    (20th root) 1.25=r
    r=1.01

    200000=100000(1.01)^t
    2=1.01^t
    log 2/log1.01=t
    t=62 minutes which is the answer in the back of my textbook. Thanks for setting me on the right path Char.Limit :D
     
  8. Jan 25, 2010 #7

    Char. Limit

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    No problem. And it's true... roots are almost always easier than logs.
     
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