Formula
The exponential equation is `y=ab^x`
taking logarithm on both sides, we get
`log_(10)y=log_(10)(ab^x)`
`log_(10)y=log_(10)a+log_(10)(b^x)`
`log_(10)y=log_(10)a+x log_(10)b`
`Y=A+Bx` where `Y=log_(10)y, A=log_(10)a, B=log_(10)b`
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=ab^x)` - Curve fitting using Least square method
| X | Y |
| 0 | 10 |
| 1 | 21 |
| 2 | 35 |
| 3 | 59 |
| 4 | 92 |
| 5 | 200 |
| 6 | 400 |
| 7 | 610 |
Solution:
The curve to be fitted is `y=ab^x`
taking logarithm on both sides, we get
`log_(10)y=log_(10)a+x log_(10)b`
`Y=A+Bx` where `Y=log_(10)y, A=log_(10)a, B=log_(10)b`
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 | 10 | 1 | 0 | 0 |
| 1 | 21 | 1.3222 | 1 | 1.3222 |
| 2 | 35 | 1.5441 | 4 | 3.0881 |
| 3 | 59 | 1.7709 | 9 | 5.3126 |
| 4 | 92 | 1.9638 | 16 | 7.8552 |
| 5 | 200 | 2.301 | 25 | 11.5051 |
| 6 | 400 | 2.6021 | 36 | 15.6124 |
| 7 | 610 | 2.7853 | 49 | 19.4973 |
| --- | --- | --- | --- | --- |
| `sum x=28` | `sum y=1427` | `sum Y=15.2893` | `sum x^2=140` | `sum x*Y=64.1929` |
Substituting these values in the normal equations
`8A+28B=15.2893`
`28A+140B=64.1929`
Solving these two equations using Elimination method,
`8a+28b=15.2893`
`:.8a+28b=15.29`
and `28a+140b=64.1929`
`:.28a+140b=64.19`
`8a+28b=15.2893 ->(1)`
`28a+140b=64.1929 ->(2)`
equation`(1) xx 7 =>56a+196b=107.0251`
equation`(2) xx 2 =>56a+280b=128.3858`
Substracting `=>-84b=-21.3607`
`=>84b=21.3607`
`=>b=21.3607/84`
`=>b=0.254294`
Putting `b=0.254294` in equation `(1)`, we have
`8a+28(0.254294)=15.2893`
`=>8a=15.2893-7.120233`
`=>8a=8.169067`
`=>a=8.169067/8`
`=>a=1.021133`
`:. a=1.021133" and "b=0.254294`
we obtain `A=1.0211,B=0.2543`
`:. a=antilog_10(A)=antilog_10(1.0211)=10.4986`
and `b=antilog_10(B)=antilog_10(0.2543)=1.7959`
Now substituting this values in the equation is `y = a b^x`, we get
`y = 10.4986 * (1.7959)^x`
This material is intended as a summary. Use your textbook for detail explanation.
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