Formula
The exponential equation is `y=ae^(bx)`
taking logarithm on both sides, we get
`log_(10)y=log_(10)(ae^(bx))`
`log_(10)y=log_(10)a+log_(10)e^(bx)`
`log_(10)y=log_(10)a+bx log_(10)e`
`Y=A+Bx` where `Y=log_(10)y, A=log_(10)a, B=blog_(10)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
`log_(10)y=log_(10)a+bx log_(10)e`
`Y=A+Bx` where `Y=log_(10)y, A=log_(10)a, B=blog_(10)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=log_(10)(y)` | `x^2` | `x*Y` |
| 0 | 0.1 | -1 | 0 | 0 |
| 0.5 | 0.45 | -0.3468 | 0.25 | -0.1734 |
| 1 | 2.15 | 0.3324 | 1 | 0.3324 |
| 1.5 | 9.15 | 0.9614 | 2.25 | 1.4421 |
| 2 | 40.35 | 1.6058 | 4 | 3.2117 |
| 2.5 | 180.75 | 2.2571 | 6.25 | 5.6427 |
| --- | --- | --- | --- | --- |
| `sum x=7.5` | `sum y=232.95` | `sum Y=3.81` | `sum x^2=13.75` | `sum x*Y=10.4556` |
Substituting these values in the normal equations
`6A+7.5B=3.81`
`7.5A+13.75B=10.4556`
Solving these two equations using Elimination method,
`6a+7.5b=3.81`
and `7.5a+13.75b=10.4556`
`:.7.5a+13.75b=10.46`
`6a+7.5b=3.81 ->(1)`
`7.5a+13.75b=10.4556 ->(2)`
equation`(1) xx 7.5 =>45a+56.25b=28.575`
equation`(2) xx 6 =>45a+82.5b=62.7336`
Substracting `=>-26.25b=-34.1586`
`=>26.25b=34.1586`
`=>b=34.1586/26.25`
`=>b=1.30128`
Putting `b=1.30128` in equation `(1)`, we have
`6a+7.5(1.30128)=3.81`
`=>6a=3.81-9.7596`
`=>6a=-5.9496`
`=>a=-5.9496/6`
`=>a=-0.9916`
`:. a=-0.9916" and "b=1.30128`
we obtain `A=-0.9916,B=1.3013`
`:. a=antilog_10(A)=antilog_10(-0.9916)=0.102`
and `b=B/log_10(e)=(1.3013)/(0.4343)=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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