PreCalc
posted by Kate .
[b]1. The problem statement, all variables and given/known data[/b]
The average growth rate of the population of a certain city is 7.5% per year. The city's population is now 22,750 people. What is the expected population in 10 years?
[b]2. Relevant equations[/b]
I was always taught this formula for exponential growth
N(t) = N_0 e^(kt)
N = population
N_0 = population at t(0)
e = 2.7...
k = some positive constant
t = time
Here's what my teacher wrote on my paper for the formula
f(x) = C(1 + r)^x
f(x) = population
C = initial population
r = growth rate
[b]3. The attempt at a solution[/b]
no i don't understand how to do this exactly because I don't know what to use for the constant k
so i used the second one
22750 (1 + .075)^10 = 46888.4680
now what I don't udnerstand is that this really makes no sense at all becasue if I wanted to fidn the population at 10 minutes or ten centuries and just plugged in 10 into the equation with no units at all I would get the same exact answer. Can youp please tell me how to go about reasoning this out... THANKS!

When your teacher wrote ...
f(x) = C(1 + r)^x
f(x) = population
C = initial population
r = growth rate
he/she should have also defined x to be the annual rate.
so when you replaced x with 10 it was understood that it was years, since the r was the rate per year
If you wanted to find out for a time other than years, you would have to change t to that fraction of a year.
e.g. if you only wanted it for 10 months, t = 10/12 or .83333..
if you wanted 10 minutes you would have to find the number of minutes in a year
which is 365*24*60 = 525600
so your exponent for 10 minutes would not be 10 but
10/525600
the second equation is probably the easier to use for these types of questions.
you could use the first one
N(t) = N_{0} e^(kt)
here we have to find the value of k first by using one set of data given, that is
when t = 1,
22750(1.075) = 22750e^1k)
e^k = 1.075
k = ln 1.075 = .07232
so N(t) = 22750e^.07232t
so when t=10
N(10) = 22750(e^(.07232)(10))
= 46888.46804 exactly the same as obtained with the other formula.