1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Exponential Functions

  1. Jun 21, 2016 #1
    1. The problem statement, all variables and given/known data
    In 2003 the city of spring field had a population of 250000 the population is expected to double by 2025, how many people in 2015?

    2. Relevant equations

    3. The attempt at a solution

    The initial is 250000 and b is 2 because it doubles however I am unsure of what the exponent is I've tried but I can't get it. Also don't you have to put 500000 in A because we already know the final amount? Help is appreciated thanks.
  2. jcsd
  3. Jun 21, 2016 #2

    Buzz Bloom

    User Avatar
    Gold Member

    Hi Veronica:

    I think that the equation A=Pb^t is not going to help you.
    Try instead A = P b^(t-2003).
    This equation makes it easier to calculate P.

    I suggest you organize the data as a table like the following:
    t=Year A=Population b^(t-2003)
    1) 2003 P=250000 b^0=1
    2) 2015 X=??? b^(2015-2003)=???
    3) 2025 500000 b^(2025)=2​

    Use (3) to find b. Then using this value for b, use the equation to find b^(2015-2013) and X.

    Hope this helps,
  4. Jun 21, 2016 #3
    The rate of growth can be calculated by the given data and then the population after a time span can be found if the rate is taken as steady/uniform.
    PR = N(T2)- N(T1)/ (T2-T1) = dN/dT
  5. Jun 22, 2016 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Probably not too useful in the precalculus Forum.
  6. Jun 22, 2016 #5


    User Avatar
    Homework Helper

    Like Buzz pointed out, you could fix t to years, i.e. (year - 2013) and change your b. Otherwise you could fix your b and change your time. In either case, you need to achieve the goal of defining this as a function of the year and then solve for the year 2015.

    If you fix b = 2 and change your exponent, you need to define your exponent to be equal to 0 in 2013 and 1 in 2025.
    This way A = Pb^t = 250000* 2^t will give you 250000 in year 2013 and 500000 in 2025.

    Can you think of a way to do that?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted