Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Poplulation growth problem

  1. Jul 20, 2006 #1
    im trying to solve a population growth problem. not sure if im using the right eqn. Please check:

    Q: if the population doubles in two years, how long does it take to triple?

    Solution:

    general equation for population growth: N=No*e^(rt)
    1) find unknown constant "r" from given
    2No=No*e^(rt)
    from equation, r= ln(2)/2
    2) find "t" for population to triple
    3No=No*e^(rt)
    solving for t: t=2ln(3)/ln(2) ~ 3.170 years
     
  2. jcsd
  3. Jul 20, 2006 #2
    If thats the equation your given then yes you are using the right equation. Just verify that t is indeed measured in years rather than some other time scale. Also verify that they imply triple to mean 3No not 3(2No). Even thoughy you used a calculator i think you can still simplify ln(3)/ln(2)
     
  4. Jul 20, 2006 #3
    thats the problem im not sure im using the right equation.
     
  5. Jul 21, 2006 #4

    HallsofIvy

    User Avatar
    Science Advisor

    Yes, that's a perfectly valid formula.

    However, you could also use
    [tex]P(t)= P_0 2^{\frac{t}{2}}[/tex]
    (every two years, t/2 is an integer, so we have multiplies by 2 t/2 times.)
    then
    [tex]P(t)= 3P_0= P_0 2^{\frac{T}{2}[/tex]
    [tex]2^{\frac{T}{2}}= 3[/tex]
    [tex]\left(\frac{T}{2}\right) log(2)= log(3)[/tex]
    [tex]T= \frac{2 log(3)}{ log(2)}[/tex]
    as you have.
     
  6. Jul 21, 2006 #5
    also note that your equation when substituted with your given value of "r" simplies to the equation posted by HallsOfIvy
     
  7. Jul 21, 2006 #6

    HallsofIvy

    User Avatar
    Science Advisor

    Exactly. All "exponentials" are interchangable. That's why you only need log base 10 and log base e on your calculator.
    [tex]2^{\frac{t}{2}}= e^{ln(2^{\frac{t}{2}}= e^\frac{t}{2}ln(2)[/tex]
    which is [itex]e^{rt}[/itex] with r= ln(2)/2.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook