1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Staff Emeritus
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Exponential population decay word problem
  1. Exponential Decay (Replies: 8)

Loading...