# Exponential Functions

1. Jun 21, 2016

### Veronica_Oles

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
A=Pb^t

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. Jun 21, 2016

### Buzz Bloom

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,
Buzz

3. Jun 21, 2016

### drvrm

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

4. Jun 22, 2016

### SammyS

Staff Emeritus
Probably not too useful in the precalculus Forum.

5. Jun 22, 2016

### RUber

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?