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

x 1 2 3 4
y 1.65 2.70 4.50 7.35

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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:

iteration xi yi log yi xlog yi xi2
1 1 1.65 0.5 0.825 1
2 2 2.7 0.99 2.673 4
3 3 4.5 1.504 6.768 9
4 4 7.35 1.99 14.6265 16
n =4 \(\sum x_i = 10\) \(\sum log y_i = 4.984\) \(\sum x_i \enspace logy_i = 24.8925\) \(\sum x_i^2 = 30\)

Now,

\(b = \frac{4 \times (24.892) – 10\times (4.984)}{4 \times 30 – 10^2}\)

b = 2.4865

\(log a = \frac{4.984}{4} – (2.4865)\frac{10}{4}\)

log a = -4.97025

a = e-4.97025 = 0.00694

Thus regression formula is

y = 0.00694 x e-2.4865x

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