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Homework Help: Exponential population decay word problem

  1. Jan 2, 2006 #1
    Suppose that in any given year, the population of a certain endangered species is reduced by 25%. If the population is now 7500, in how many years will the population be 4000?
    Well, I know that the formula for this is P(t) = Cekt.
    I just can't figure out the value of k.
    i did 7500 x .25 = 1875... 7500 - 1875 = 5625...
    so i set up 7500x = 5625 and got .75.
    so, essentially.. the population is growing at .75 per year.
    so i used .75 as k and:
    4000 = 7500e^(.75t)
    4000/7500 = e^(.75t)
    8/15 = e^(.75t)
    ln(8/15) = .75t lne
    ln(8/15) = .75t
    t = [ln(8/15)] / .75
    t = -.8381448792 years
    This can't be correct because time isn't negative, and if you do it logically, it would be around 2. something years:
    7500 x .75 = 5625 ----> 1 year
    5625 x .75 = 4218.75 ----> 2 yrs
    4218.75 x .75 = 3164.0625 ----> 3 yrs
    So it's somewhere between 2 and 3 years... closer to 2 years.
    and if i set up 4218.75x = 4000
    x = .9481481481
    so i know after 2 years.. i can multiply by .9481481481 and i get 4000..
    but i don't know how to convert that into years.. since that is a percentage.
    BLEH.. brain is about to explode.. don't know what to do.. lol :bugeye:
     
  2. jcsd
  3. Jan 2, 2006 #2
    Should K be -.25? The time comes out to 2.514 then
     
  4. Jan 2, 2006 #3

    HallsofIvy

    User Avatar
    Science Advisor

    There is no reason at all to think that k= 0.75! Don't just plug numbers in at random- think!
    Each year, the population decreases by 0.25 and so must be 0.75 times the previous years population. That is, if this year's (t= 1) population is P, next year's is 0.75P: 0.75P= Pek or ek= 0.75. Solve that for k.

    Personally, I wouldn't use "ekt" at all. The population is multiplied by 0.75 each year so its population after t years would be P0(0.75)t. You want to solve
    7500(0.75)t= 4000.
     
  5. Jan 2, 2006 #4
    got it.. 2.1850811 years
     
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