1. Find Solution using Lagrange's Interpolation formula

x	f(x)
2	0.69315
2.5	0.91629
3	1.09861

x = 2.7

Solution:
The value of table for `x` and `y`

x	2	2.5	3
y	0.69315	0.91629	1.09861

Lagrange's Interpolating Polynomial
The value of x at you want to find `P_n(x) : x = 2.7`

Lagrange's formula is
`f(x) = ((x - x_1)(x - x_2))/((x_0 - x_1)(x_0 - x_2)) xx y_0 + ((x - x_0)(x - x_2))/((x_1 - x_0)(x_1 - x_2)) xx y_1 + ((x - x_0)(x - x_1))/((x_2 - x_0)(x_2 - x_1)) xx y_2`

`y(2.7) = ((2.7 - 2.5)(2.7 - 3))/((2 - 2.5)(2 - 3)) xx 0.69315 + ((2.7 - 2)(2.7 - 3))/((2.5 - 2)(2.5 - 3)) xx 0.91629 + ((2.7 - 2)(2.7 - 2.5))/((3 - 2)(3 - 2.5)) xx 1.09861`

`y(2.7) = ((0.2)(-0.3))/((-0.5)(-1)) xx 0.69315 + ((0.7)(-0.3))/((0.5)(-0.5)) xx 0.91629 + ((0.7)(0.2))/((1)(0.5)) xx 1.09861`

`y(2.7) = (-0.06)/(0.5) xx 0.69315 + (-0.21)/(-0.25) xx 0.91629 + (0.14)/(0.5) xx 1.09861`

`y(2.7) = 0.994116`


Solution of the polynomial at point `2.7` is `y(2.7) = 0.994116`

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