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  • Please solve RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 9 sub question (viii) maths textbook solution

Please solve RD Sharma class 12 chapter Differentials Errors and Approximations exercise 13.1 question 9 sub question (viii) maths textbook solution

Answers (1)

Answer:  1.396368

Hint:  Here, we use the formula

        \Delta y=f(x+\Delta x)-f(x)

Given:  \log _{10} 4=0.6021

        \log _{10} \mathrm{e}=0.4343

Solution: 

consider  \mathrm{y}=\log _{10} x

Or   \mathrm{y}=\frac{\log _{10} \mathrm{x}}{\log _{10} \mathrm{e}}

We get

        y=0.4343 \log _{e} x

Here  

        \mathrm{x}=4 \text { and } \Delta \mathrm{x}=.04

On differentiating wrt x

        \begin{aligned} &\frac{d y}{d x}=\frac{0.4343}{x} \Delta \mathrm{x} \\\\ &\text { At } \mathrm{x}=4 \\\\ &\frac{d y}{d x}=\frac{1}{4} \end{aligned}

\begin{aligned} &\text { So }, \Delta y=\frac{d y}{d x} d x=\frac{1}{4} 0.04=0.01 \\\\ &\text { i.e } \Delta y=0.01 \\\\ &\text { Therefore } \log _{\mathrm{e}} 4.04=\mathrm{y}+\Delta \mathrm{y}=1.396368 \end{aligned}

 

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