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Math Help - Exponential Function

  1. #1
    mathnerd1993
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    Exclamation 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!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathnerd1993 View Post
    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
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  3. #3
    Global Moderator

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    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.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    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
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