Formula
The exponential equation is `y=ax^b`
taking logarithm on both sides, we get
`log_(10)y=log_(10)(ax^b)`
`log_(10)y=log_(10)a+log_(10)(x^b)`
`log_(10)y=log_(10)a+b log_(10)x`
`Y=A+bX` where `Y=log_(10)y, A=log_(10)a, X=log_(10)x`
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=ax^b)` - Curve fitting using Least square method
Solution:
The curve to be fitted is `y=ax^b`
taking logarithm on both sides, we get
`log_(10)y=log_(10)a+b log_(10)x`
`Y=A+bX` where `Y=log_(10)y, A=log_(10)a, X=log_(10)x`
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` | `X=log_(10)(x)` | `Y=log_(10)(y)` | `X^2` | `X*Y` |
| 2 | 27.8 | 0.301 | 1.444 | 0.0906 | 0.4347 |
| 3 | 62.1 | 0.4771 | 1.7931 | 0.2276 | 0.8555 |
| 4 | 110 | 0.6021 | 2.0414 | 0.3625 | 1.229 |
| 5 | 161 | 0.699 | 2.2068 | 0.4886 | 1.5425 |
| --- | --- | --- | --- | --- | --- |
| `sum x=14` | `sum y=360.9` | `sum X=2.0792` | `sum Y=7.4854` | `sum X^2=1.1693` | `sum X*Y=4.0618` |
Substituting these values in the normal equations
`4A+2.0792b=7.4854`
`2.0792A+1.1693b=4.0618`
Solving these two equations using Elimination method,
`4a+2.0792b=7.4854`
`:.4a+2.08b=7.49`
and `2.0792a+1.1693b=4.0618`
`:.2.08a+1.17b=4.06`
`4a+2.0792b=7.4854 ->(1)`
`2.0792a+1.1693b=4.0618 ->(2)`
equation`(1) xx 1.1693 =>4.6772a+2.431209b=8.752678`
equation`(2) xx 2.0792 =>4.323073a+2.431209b=8.445295`
Substracting `=>0.354127a=0.307384`
`=>a=0.307384/0.354127`
`=>a=0.868003`
Putting `a=0.868003` in equation `(2)`, we have
`2.0792(0.868003)+1.1693b=4.0618`
`=>1.1693b=4.0618-1.804752`
`=>1.1693b=2.257048`
`=>b=2.257048/1.1693`
`=>b=1.930256`
`:. a=0.868003" and "b=1.930256`
we obtain `A=0.868,b=1.9303`
`:. a=antilog_10(A)=antilog_10(0.868)=7.3791`
Now substituting this values in the equation is `y = a x^b`, we get
`y = 7.3791 * x^(1.9303)`
This material is intended as a summary. Use your textbook for detail explanation.
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