The approximate value of $\log _{10} 1002$ is (Given $\log _{10} \mathrm{e}=0.4343$ )

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } \mathrm{f}(x)=\log _e x \ & 	herefore \quad \mathrm{f}^{\prime}(x)=\frac{1}{x} \ & 	ext { Here, } a=1000, \mathrm{~h}=2 \ & 	herefore \quad \mathrm{f}(\mathrm{a}+\mathrm{h}) \approx \mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a}) \ & \approx f(1000)+2 f^{\prime}(1000) \ & \approx \log _{\mathrm{e}} 10^3+2 	imes \frac{1}{1000} \ & \approx 3(0.4343)+0.002 \ & \approx 1.3029+0.002 \ & \approx 1.3049 \ & 	herefore \quad \log _{\mathrm{e}} 1002 \approx 1.3049 \ & 	ext { Now, } \log _{10} 1002 \approx \frac{\log _{\mathrm{e}} 1002}{\log _{\mathrm{e}} 10} \ & \approx \frac{1.3049}{0.4343} \approx 3.0009 \end{aligned}$

The approximate value of $\log _{10} 1002$ is (Given $\log _{10} \mathrm{e}=0.4343$ )

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