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`