Question

Certain corresponding values of x and log10x are given below:

x: 300 304 305 307

log10 x: 2.4771 2.4829 2.4843 2.4871

Find log10 (310) by Lagrange’s formula.

Answer 1

Here the intervals are unequal.

$$\begin{matrix} x_0=300 & y_0=2.4771 \\ \\ x_1=304 & y_1=2.4829 \\ \\ x_2=305 & y_2=2.4843 \\ \\ x_3=307 & y_3=2.4871 \\ \end{matrix}$$

By Lagrange’s interpolation formula we have

$$y=f(x)=\dfrac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\times y_0$$

$$+\dfrac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\times y_1$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\times y_2$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\times y_3$$

Put $x=310$

$$f(310)=\dfrac{(310-304)(310-305)(310-307)}{(300-304)(300-305)(300-307)}\times 2.4771$$

$$+\dfrac{(310-300)(310-305)(310-307)}{(304-300)(304-305)(304-307)}\times 2.4829$$

$$+\dfrac{(310-300)(310-304)(310-307)}{(305-300)(305-304)(305-307)}\times 2.4843$$

$$+\dfrac{(310-300)(310-304)(310-305)}{(307-300)(307-304)(307-305)}\times 2.4871$$

$$=2.4914$$

$\log_{10}(310)=2.4914$