Question

Using Newton-Raphson method to find the approximate root of the equation x log_{10}x=1.2

( Carry out three iterations ).
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Thank uh!

Answer 1

`x \approx 2.740646096`

Explanation:

Given:

`x\log_(10)x=1.2`

Rewrite so that it's equal to zero:

`\implies x\log_(10)x-1.2=0`

Therefore:

`\implies f(x)=x\log_(10)(x)-1.2`

Differentiate the function:

Use the product rule to differentiate
`x\log_(10)(x)`:

`\textsf{Let }u=x \implies (du)/(dx)=1`

`\textsf{Let }v=\log_(10)(x) \implies (dv)/(dx)=(1)/(x \ln 10)`

`\begin{aligned}\implies (dy)/(dx) & = u(dv)/(dx)+v (du)/(dx)\\& = (x)/(x \ln 10)+\log_(10)(x)\\& = (1)/(\ln 10)+\log_(10)(x)\end{aligned}`

`\implies f'(x)=(1)/(\ln 10)+\log_(10)(x)`

`\implies f'(x)=(1+\ln 10 \log_(10)(x))/(\ln 10)`

Newton-Rhapson iteration formula

`x_(n+1)=x_n-(f(x_n))/(f'(x_n))`

Substitute the function and its derivative into the formula, replacing x with
`x_n`:

`\implies x_(n+1)=x_n-(x_n\log_(10)(x_n)-1.2)/((1+\ln 10 \log_(10)(x))/(\ln 10))`

`\implies x_(n+1)=x_n-(x_n\ln 10\log_(10)(x_n)-1.2\ln 10)/(1+\ln 10 \log_(10)(x_n))`

Therefore, the iteration formula is:

`x_(n+1)=x_n-(x_n\ln 10\log_(10)(x_n)-1.2\ln 10)/(1+\ln 10 \log_(10)(x_n))`

Let
`x_0=3`

Substitute this into the formula and carry out 3 iterations:

`\implies x_(1)=3-(3\ln 10\log_(10)(3)-1.2\ln 10)/(1+\ln 10 \log_(10)(3))=2.746149035`

`\implies x_2=2.740648841`

`\implies x_3= 2.740646096`