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)`