Formula
The exponential equation is `y=ae^(bx)`
taking natural logarithm on both sides, we get
`ln(y)=ln(ae^(bx))`
`ln(y)=ln(a)+ln(e^(bx))`
`ln(y)=ln(a)+bx ln(e)`
`Y=A+Bx` where `Y=ln(y), A=ln(a), B=bln(e)`
which linear in Y,x
So the corresponding normal equations are
`sum Y = nA + B sum x`
`sum xY = A sum x + B sum x^2`
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Examples
1. Calculate Fitting exponential equation `(y=ae^bx)` - Curve fitting using Least square method
| X | Y |
| 0 | 0.10 |
| 0.5 | 0.45 |
| 1 | 2.15 |
| 1.5 | 9.15 |
| 2 | 40.35 |
| 2.5 | 180.75 |
Solution:
The curve to be fitted is `y=ae^(bx)`
taking logarithm on both sides, we get
`ln(y)=ln(a)+bx ln(e)`
`Y=A+Bx` where `Y=ln(y), A=ln(a), B=bln(e)`
which linear in Y,x
So the corresponding normal equations are
`sum Y = nA + B sum x`
`sum xY = A sum x + B sum x^2`
The values are calculated using the following table
| `x` | `y` | `Y=ln(y)` | `x^2` | `x*Y` |
| 0 | 0.1 | -2.3026 | 0 | 0 |
| 0.5 | 0.45 | -0.7985 | 0.25 | -0.3993 |
| 1 | 2.15 | 0.7655 | 1 | 0.7655 |
| 1.5 | 9.15 | 2.2138 | 2.25 | 3.3206 |
| 2 | 40.35 | 3.6976 | 4 | 7.3952 |
| 2.5 | 180.75 | 5.1971 | 6.25 | 12.9928 |
| --- | --- | --- | --- | --- |
| `sum x=7.5` | `sum y=232.95` | `sum Y=8.7728` | `sum x^2=13.75` | `sum x*Y=24.0748` |
Substituting these values in the normal equations
`6A+7.5B=8.7728`
`7.5A+13.75B=24.0748`
Solving these two equations using Elimination method,
`6a+7.5b=8.7728`
and `7.5a+13.75b=24.0748`
`:.7.5a+13.75b=24.07`
`6a+7.5b=8.7728 ->(1)`
`7.5a+13.75b=24.0748 ->(2)`
equation`(1) xx 7.5 =>45a+56.25b=65.796`
equation`(2) xx 6 =>45a+82.5b=144.4488`
Substracting `=>-26.25b=-78.6528`
`=>26.25b=78.6528`
`=>b=78.6528/26.25`
`=>b=2.996297`
Putting `b=2.996297` in equation `(1)`, we have
`6a+7.5(2.996297)=8.7728`
`=>6a=8.7728-22.472229`
`=>6a=-13.699429`
`=>a=-13.699429/6`
`=>a=-2.283238`
`:.a=-2.283238" and "b=2.996297`
we obtain `A=-2.2832,B=2.9963`
`:. a=antiln(A)=antiln(-2.2832)=0.102`
and `b=B/ln(e)=(2.9963)/(1)=2.9963`
Now substituting this values in the equation is `y = a e^(bx)`, we get
`y=0.102*e^(2.9963x)`
This material is intended as a summary. Use your textbook for detail explanation.
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