We will first create a table of data pairs (x,lny) using the given information.
Age (in years), x
|
20
|
30
|
40
|
50
|
60
|
Near point (in centimeters), y
|
12
|
15
|
25
|
40
|
100
|
lny
|
2.48
|
2.71
|
3.22
|
3.69
|
4.61
|
Next, we will plot the transformed points.
The points lie close to a line, so an exponential model should be a good fit for the original data. To write an exponential model
y=abx, let's use the points
(30,2.71) and
(60,4.61) because they appear to be on the line. With this, we will write an equation in
lny−lny1=m(x−x1)
In this form,
m is the and
(x1,lny1) is a reference point that is on the line. Using the we can substitute the chosen points and find the slope.
m=x2−x1y2−y1
m=60−304.61−2.71
m=301.9
m=0.063333…
m≈0.06
Now that we found the slope, we can write the equation in point-slope form. Let the point
(30,2.71) be our reference point.
lny−2.71=0.06(x−30)
Next, we will isolate
y to have the model in the form of
y=abx.
lny−2.71=0.06(x−30)
lny−2.71=0.06x−1.8
lny=0.06x+0.91
elny=e0.06x+0.91
y=e0.06x+0.91
y=e0.06x⋅e0.91
y=(e0.06)x⋅e0.91
y=(1.06183…)x⋅2.48432…
y=2.48432…(1.06183…)x
y=2.48(1.06)x
As a result, one possible model is
y=2.48(1.06)x depending on the choice of the points.