Thread: Exponential Function

Over past 2 decades, # of computers in schools has been increasing. The data shows the # of students per computer in U.S public schools. Assume that eventually, there will be 1 student per computer.

School year=Students per computer
1983-84= 125
1984-85= 75
1985-86= 50
1986-87= 37
1987-88= 32
1988-89= 25
1989-90= 22
1990-91= 20
1991-92= 18
1992-93= 16
1993-94= 14
1994-95= 10.5
1995-96= 10
1996-97= 7.8
1997-98= 6.1
1998-99= 5.7
1999-2000= 5.4

a. Find an exponential function in the form y=abxthat models the data. Let x be the number of school years after the 1983-84 school year.

Here is how I worked the problem:
y=y b(x-x)
(x,y)=(1, 75)
(x,x)=(0, 125)
75=125b1-0
.6=b
y=125(.6)x

Did I solve this question right?
~thanks!

Originally Posted by mathnerd1993

Over past 2 decades, # of computers in schools has been increasing. The data shows the # of students per computer in U.S public schools. Assume that eventually, there will be 1 student per computer.

School year=Students per computer
1983-84= 125
1984-85= 75
1985-86= 50
1986-87= 37
1987-88= 32
1988-89= 25
1989-90= 22
1990-91= 20
1991-92= 18
1992-93= 16
1993-94= 14
1994-95= 10.5
1995-96= 10
1996-97= 7.8
1997-98= 6.1
1998-99= 5.7
1999-2000= 5.4

a. Find an exponential function in the form y=abxthat models the data. Let x be the number of school years after the 1983-84 school year.

Here is how I worked the problem:
y=y b(x-x)
(x,y)=(1, 75)
(x,x)=(0, 125)
75=125b1-0
.6=b
y=125(.6)x

Did I solve this question right?
~thanks!

A model y=abx is a linear model, not exponential. An exponential model would be of the form or . I would recommend doing a regression fit using the second form.

Exponential regression can be done on a calculator, if the calculator has the correct package. But we can "massage" the data a little to get it into a form where we can do a linear regression to get the coefficients a and b.

Given the equation , take the natural logartithm of both sides:


or


This is now in the form: (a line) where , , and

So do the following to your data set: (I am setting the year '83 - '84 to be x = 1)
1, ln(125)
2, ln(75)
3, ln(50)
.
.
.

Do a linear regression on this data. You'll get values M for the slope and B for the intercept.

Then your value for a will be and your value for b will be .

(I got a = 90.93000 and b = -0.17638 using the model . The correlation coefficient was 0.96 or so, indicating a good fit.)

-Dan

I assume you need to use least squares to do this. I do not think the homework was intended by hand. I used computer software to get,

The when I told the program to find a power approximation it said "Abnormal Program Termination" and it crashed. I lost the picture thus you will not be able to see it. I believe when I does the solution it takes logarithsm and a negative logarithm appears somewhere which causes it to crash.
Where is the amount of years passed.

Originally Posted by ThePerfectHacker

I assume you need to use least squares to do this. I do not think the homework was intended by hand. I used computer software to get,

The when I told the program to find a power approximation it said "Abnormal Program Termination" and it crashed. I lost the picture thus you will not be able to see it. I believe when I does the solution it takes logarithsm and a negative logarithm appears somewhere which causes it to crash.
Where is the amount of years passed.

Taking logs of the number of computers will help in that linear regression
can then be used which can be done by hand or with a spreadsheet.

The problem is that a power law is incompatible with the number of
computers per student tending to 1.

Also plotting best fit curves indicates that the data do not fit a simple
power law very well.

There are other objections to this question but I think these two will do
for a start.

RonL