# A Point on the Best-Fit Curve

All exponential equations are of the form.

### y = a e^(bx)

For our problem we have to find the a and b in the equation that best fits the data.

The first step in an exponential fit analysis is to find one point that must lie on the best fitting curve to the data. To do this we need to do a linear transformation to the population data. Here's how.

An initial guess for this point might be the mean data point since this point would fall on the best fitting line in a linear analysis.

In the spreadsheet, double click on the graph.

• Once the graph is selected, double click on the y-axis.
• In the corner there is an option called Logarthimic Scale. Click on its box.
• Click O.K.

What do you notice about the graph? Does it look more like a straight line?

What happened was that the population data was plotted on a lograthimic scale where its values have been changed to its natural log. Since this looks like a straight line, we can use the mean data point of this graph as the best-fitting point.

In the spreadsheet, find the natural log of the population data (column D).
Find the sum (cell D23) and the mean (cell D24) of this list.

This point fits well to our data since the population data has been transformed by taking the natural log. How can we "untransform" this point?

Well, since the data was transformed by the natural log, the inverse of the natural log is the exponential. So, if we find the exponential of the mean, we can "untransform" this point to our actual data.

Do this in the spreadsheet in cell D25.

Look at this plot.

This is a plot of the years and regular population data and the point is the values from cells B24 and D26.

Do you think this point fits the best fitting curve reasonably well?

In summary, to find a point that must lie on the best fitting curve you must:
1. Take the natural log of each y-value data point.
2. Find the mean of these points.
3. Raise e to the number you found in 2).

We can also show that this point must be on the best fitting curve analytically. Click here to see how.

## Exercises

1.) Decide if the following plots of data should be represented by an exponential fit or a linear fit and explain why!

a.) This is the plot of the Dow Jones Industrial Average for every third year from 1930 to 1990. The year 1930 is designated year 0.

This data was found on the Internet at http://www.geom.umn.edu/docs/snell/chance/teaching_aids/data_sets.html It contains the Dow Jones Average on the first trading day of the year from 1930-1992.

b.) This is a plot of the diameters and circumference of various objects.

c.) A radioactive element is one that is unstable and decays into other more stable elements. This is a plot of the amounts of Polonium, a radioactive element, at times recorded every 50 days.

2.) Take a look at the Dow Jones data from the previous question.

DOW JONES INDUSTRIAL AVERAGE
TIME (years 0=1930)AVERAGE
0244.2
359.29
6144.13
9153.64
12112.77
15152.58
18181.04
21239.92
24282.89
27490.03
DOW JONES INDUSTRIAL AVERAGE
TIME (years 0=1930)AVERAGE
30679.06
33646.79
36968.54
39947.73
42809.30
45632.04
48817.74
51972.78
541252.74
571927.31
602810.15

Find the point that must fall on the best fitting curve of this data.

3.) Here is the data for the radioactive element.

POLONIUM DATA
TIME (days)AMOUNT (mg)
0100
5076
10062
15047
20037
25029
30021
35017