Fit the curve \(y = ae^{bx}\) through the following data points.

x 0 1 3 5 7 9
y 1.0 0.891 0.708 0.563 0.447 0.355

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Solution:

We know that exponential model is given by

y = aebx

Taking log on both side, We get

log y = log a + bx

This equation is similar in form to the linear equation y = a + bx. Thus we evaluate parameters a and b by using the equation of linear regression model as below:

\(b = \frac{n\sum_{i=1}^{n} x_i log y_i – \sum_{i=1}^{n}x_i \sum_{i=1}^{n} log y_i}{n\sum_{i=1}^{n}x_i^2 – (\sum_{i=1}^{n}x_i)^2}\)

And

\(loga = \frac{\sum_{i=1}^n log y_i}{n}-b\frac{\sum_{i=1}^n x_i}{n}\)

Now we calculate required summations as below:

i xi yi log yi xlog yi xi2
1 0 1 0.0000 0.0000 0.00
2 1 0.891 -0.11541 -0.1154 1.00
3 3 0.708 -0.34531 -1.0359 9.00
4 5 0.562 -0.57625 -2.8813 25.0
5 7 0.447 -0.80520 -5.6364 49.0
6 9 0.355 -1.0356 -9.3207 81.0
n =6 \(\sum x_i = 25\) \(\sum log y_i = -2.878\) \(\sum x_i \enspace logy_i = -18.99\) \(\sum x_i^2 = 165\)

Now,

\(b = \frac{6 \times (-18.99) – 25 \times (-2.878)}{6 \times 165 – 25^2}\)

b = -0.115

\(log a = \frac{-2.878}{6} – (-0.115)\frac{25}{6}\)

log a = -0.0005

a = e-0.0005 = 0.999

Thus regression formula is

y = 0.999 x e-0.115x

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