Question

Find the root of equation x log10(x) = 1.2 by Secant method correct upto three places of decimals.​

Answer 1

Answer:

Suppose f(x)=x log10 x-1.2

Taking,

And

f(2)=2log10 (2)-12=-0.59794(-ve) f(3) =3log10 (3)-1.2=0.23136(ve)

Therefore, the root lies between 2 and 3.

1st Approximation:

2+3 2 <= 2.5

ƒ(2.5)=2.5log10 (2.5)-1.2--0.20514(-ve)

Roots lie between 2.5 and 3.

and

A

2nd Approximation

25+3 2 x₂ = = 2.75

and f(2.75)=2.75logio (2.75)-1.2-0.00816(+ve) Roots lie between 2.5 and 2.75.

3rd Approximation http://www.rgpvonline.com

2.5+2.75 2 x3 = -= 2.625 ƒ(2.625) = 2.625 log10 (2.625)-1.2=-0.09978 (-ve)

Roots lie between 2.625 and 2.75.

and

Approximation 2.625 +2.75 2 X4 = = 2.6875

and

and

= 2.74218

(2.6875)= 2.6875 log10 (2.6875)-1.2=-0.04612 (-ve) Roots lie between 2.6875 and 2.75.

5th Approximation

2.6875+2.75 2 x₂ = = 2.71875

and

f(2.71875)=2.71875 log10 (2.71875)-1.2--0.01905(-ve) Roots lie between 2.71875 and 2.75.

6th Approximation

2.71875+2.75 2 x6 = = 2.734375

ƒ(2.734375)=2.734375 log10 (2.734375) -1.2-0.005466(-ve)

Roots lie between 2.71875 and 2.75.

7th Approximation

2.734375+2.75

2

X7 =