# Exponential function help

• Dec 10th 2013, 12:40 PM
John108
Exponential function help
U.S.Department estimatedthat there were 54 million Internet users in the country in 1999 and 85 millionusers in 2002.
a)
 t(years after 1995 4 17 internet users (millions) ? ?

b)Use the information on the table to obtain an exponential function that models then umber of Internet users as a function of the time, U(t)=U0e^kt
c)rate of increase of the number of internet users in the US?
d) number of Internet users in the US in 2013 according to your model?
e)
If the population of the U.S. is approximately 314 million, for how long is this model likely to remain accurate?
new to this having problems trying to solve it
• Dec 10th 2013, 12:47 PM
romsek
Re: Exponential function help
part (a) isn't understandable as posted. Could you repost this and fix the formatting?

Have you tried any of this? Do you have any work so far you can post?
• Dec 10th 2013, 12:58 PM
John108
Re: Exponential function help
okay fixed it. im new to this and having issues trying to approach and solve this problem
• Dec 10th 2013, 01:04 PM
romsek
Re: Exponential function help
the data on #internet users came through as ?'s

I'm going to guess you want to model the data as a a scaled exponential function

N = A eb(t-t0)

where
N = #internet users
t = current year
t0 = some reference year. This is optional but useful. You specify this.

A and b are parameters to be solved for.

If you take the log of both sides you get

ln(N) = ln(A) + b(t-t0)

This is now just a line in t with slope b and value ln(A) at t=t0.

You can do a linear regression using t0 and data samples of t and ln(N) to get least square estimates of ln(A) and b.

then plug those parameter estimates into your exponential formula for N to get the estimate of N given some year t.
• Dec 10th 2013, 01:09 PM
John108
Re: Exponential function help
i need to figure out what would go there
• Dec 10th 2013, 01:20 PM
romsek
Re: Exponential function help
oh duh.. the data is in the word problem. Sorry.

basically you'd do what I just went through with
t0 = 1995
t = {1999, 2002}
N = {54, 85} in millions

now use the parameters you solved for in the exponential model to forecast for t = {1995+5, and 1995+17}
• Dec 10th 2013, 01:37 PM
John108
Re: Exponential function help
what i did for B) was u(t)=U0e^kt
y=ab^x
y=(29.5)(1.163)^x

k=ln3
k=ln(1.163)^x
k=0.151
increase at 15.1%
U(t)=29.5e^(0.151)(t)

for D) 2013 t=18
U(t)=29.5e^0.151^(18)
U(t)=446.9 million

for E) (314/29.5)= 29.5e^0.151(t) /29.5
10.64=e^0.151(t)
ln(10.64)=ln(0.151(t)e)
ln(10.64)=(0.151)(t)
ln(10.64)/0.151=t
t=15.7 years

can you check it out
• Dec 10th 2013, 02:41 PM
romsek
Re: Exponential function help
this isn't very clear.

this is what I got

Attachment 29899

first normalize the years by subtracting 1995
then take the Log of the users
rearrange the data so it's in the form {normalized year, Log(users)}

Let Mathematica find the best linear fit

Pull the coefficients out and label them as I said in my earlier post.
Convert lnA to A by taking it's exponential

Write the function that's your exponential model as I explained.
Use this function to estimates the #users in 1999, and 2012

Double check estimate for years we have data on.