Answer:
[tex]x \approx 2.740646096[/tex]
Step-by-step explanation:
Given:
[tex]x\log_{10}x=1.2[/tex]
Rewrite so that it's equal to zero:
[tex]\implies x\log_{10}x-1.2=0[/tex]
Therefore:
[tex]\implies f(x)=x\log_{10}(x)-1.2[/tex]
Differentiate the function:
Use the product rule to differentiate [tex]x\log_{10}(x)[/tex]:
[tex]\textsf{Let }u=x \implies \dfrac{du}{dx}=1[/tex]
[tex]\textsf{Let }v=\log_{10}(x) \implies \dfrac{dv}{dx}=\dfrac{1}{x \ln 10}[/tex]
[tex]\begin{aligned}\implies \dfrac{dy}{dx} & = u\dfrac{dv}{dx}+v \dfrac{du}{dx}\\& = \dfrac{x}{x \ln 10}+\log_{10}(x)\\& = \dfrac{1}{\ln 10}+\log_{10}(x)\end{aligned}[/tex]
[tex]\implies f'(x)=\dfrac{1}{\ln 10}+\log_{10}(x)[/tex]
[tex]\implies f'(x)=\dfrac{1+\ln 10 \log_{10}(x)}{\ln 10}[/tex]
Newton-Rhapson iteration formula
[tex]x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}[/tex]
Substitute the function and its derivative into the formula, replacing x with [tex]x_n[/tex]:
[tex]\implies x_{n+1}=x_n-\dfrac{x_n\log_{10}(x_n)-1.2}{\dfrac{1+\ln 10 \log_{10}(x)}{\ln 10}}[/tex]
[tex]\implies x_{n+1}=x_n-\dfrac{x_n\ln 10\log_{10}(x_n)-1.2\ln 10}{1+\ln 10 \log_{10}(x_n)}[/tex]
Therefore, the iteration formula is:
[tex]x_{n+1}=x_n-\dfrac{x_n\ln 10\log_{10}(x_n)-1.2\ln 10}{1+\ln 10 \log_{10}(x_n)}[/tex]
Let [tex]x_0=3[/tex]
Substitute this into the formula and carry out 3 iterations:
[tex]\implies x_{1}=3-\dfrac{3\ln 10\log_{10}(3)-1.2\ln 10}{1+\ln 10 \log_{10}(3)}=2.746149035[/tex]
[tex]\implies x_2=2.740648841[/tex]
[tex]\implies x_3= 2.740646096[/tex]
