1. ## 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!

2. 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 $y = ax^b$ or $y = ae^{bx}$. 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 $y = ae^{bx}$, take the natural logartithm of both sides:
$ln(y) = ln(a) + bx$

or
$ln(y) = bx + ln(a)$

This is now in the form: $Y = MX + B$ (a line) where $Y = ln(y)$, $M = b$, and $B = ln(a)$

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 $a = e^B$ and your value for b will be $b = M$.

(I got a = 90.93000 and b = -0.17638 using the model $y = ae^{bx}$. The correlation coefficient was 0.96 or so, indicating a good fit.)

-Dan

3. 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,
$y=76.22 \cdot \exp (-.1764 x)$
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 $x$ is the amount of years passed.

4. 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,
$y=76.22 \cdot \exp (-.1764 x)$
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 $x$ 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