5. Fitting exponential equation (y=ab^x) - Curve fitting Formula & Examples (taking log)

Formula

The exponential equation is

y=abx $y=ab^x$
taking logarithm on both sides, we get

log10y=log10(abx) $log_(10)y=log_(10)(ab^x)$

log10y=log10a+log10(bx) $log_(10)y=log_(10)a+log_(10)(b^x)$

log10y=log10a+xlog10b $log_(10)y=log_(10)a+x log_(10)b$

Y=A+Bx $Y=A+Bx$ where

Y=log10y,A=log10a,B=log10b $Y=log_(10)y, A=log_(10)a, B=log_(10)b$

which linear in Y,x

So the corresponding normal equations are

∑Y=nA+B∑x $sum Y = nA + B sum x$

∑xY=A∑x+B∑x2 $sum xY = A sum x + B sum x^2$

Examples
1. Calculate Fitting exponential equation (y=abx) $(y=ab^x)$ - 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=abx $y=ab^x$

taking logarithm on both sides, we get

log10y=log10a+xlog10b $log_(10)y=log_(10)a+x log_(10)b$

Y=A+Bx $Y=A+Bx$ where

Y=log10y,A=log10a,B=log10b $Y=log_(10)y, A=log_(10)a, B=log_(10)b$

which linear in Y,x
So the corresponding normal equations are

∑Y=nA+B∑x $sum Y = nA + B sum x$

∑xY=A∑x+B∑x2 $sum xY = A sum x + B sum x^2$

The values are calculated using the following table

x $x$	y $y$	Y=log10(y) $Y=log_(10)(y)$	x2 $x^2$	x⋅Y $x*Y$
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
---	---	---	---	---
∑x=28 $sum x=28$	∑y=1427 $sum y=1427$	∑Y=15.2893 $sum Y=15.2893$	∑x2=140 $sum x^2=140$	∑x⋅Y=64.1929 $sum x*Y=64.1929$

Substituting these values in the normal equations

8A+28B=15.2893 $8A+28B=15.2893$

28A+140B=64.1929 $28A+140B=64.1929$

Solving these two equations using Elimination method,

8a+28b=15.2893 $8a+28b=15.2893$

∴8a+28b=15.29 $:.8a+28b=15.29$

and

28a+140b=64.1929 $28a+140b=64.1929$

∴28a+140b=64.19 $:.28a+140b=64.19$

8a+28b=15.2893→(1) $8a+28b=15.2893 ->(1)$

28a+140b=64.1929→(2) $28a+140b=64.1929 ->(2)$

equation

(1)×7⇒56a+196b=107.0251 $(1) xx 7 =>56a+196b=107.0251$

equation

(2)×2⇒56a+280b=128.3858 $(2) xx 2 =>56a+280b=128.3858$

Substracting

⇒-84b=-21.3607 $=>-84b=-21.3607$

⇒84b=21.3607 $=>84b=21.3607$

⇒b=21.360784 $=>b=21.3607/84$

⇒b=0.254294 $=>b=0.254294$

Putting

b=0.254294 $b=0.254294$ in equation

(1) $(1)$ , we have

8a+28(0.254294)=15.2893 $8a+28(0.254294)=15.2893$

⇒8a=15.2893-7.120233 $=>8a=15.2893-7.120233$

⇒8a=8.169067 $=>8a=8.169067$

⇒a=8.1690678 $=>a=8.169067/8$

⇒a=1.021133 $=>a=1.021133$

∴a=1.021133 and b=0.254294 $:. a=1.021133" and "b=0.254294$

we obtain

A=1.0211,B=0.2543 $A=1.0211,B=0.2543$

∴a=antilog10(A)=antilog10(1.0211)=10.4986 $:. a=antilog_10(A)=antilog_10(1.0211)=10.4986$

and

b=antilog10(B)=antilog10(0.2543)=1.7959 $b=antilog_10(B)=antilog_10(0.2543)=1.7959$

Now substituting this values in the equation is

y=abx $y = a b^x$ , we get

y=10.4986⋅(1.7959)x $y = 10.4986 * (1.7959)^x$

This material is intended as a summary. Use your textbook for detail explanation.
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1. Formula & Examples (taking log)