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

\(\text {Let } \mathrm{f}(x)=\log _e x \)
\( \therefore \mathrm{f}^{\prime}(x)=\frac{1}{x} \)
\( \text {Here, } a=1000, \mathrm{~h}=2 \)
\( \therefore \mathrm{f}(\mathrm{a}+\mathrm{h}) \approx \mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a}) \)
\( \approx f(1000)+2 f^{\prime}(1000) \)
\( \approx \log _{\mathrm{e}} 10^3+2 \times \frac{1}{1000} \)
\( \approx 3(0.4343)+0.002 \)
\( \approx 1.3029+0.002 \)
\( \approx 1.3049 \)
\( \therefore \log _{\mathrm{e}} 1002 \approx 1.3049 \)
\( \text {Now, } \log _{10} 1002 \approx \frac{\log _{\mathrm{e}} 1002}{\log _{\mathrm{e}} 10} \)
\( \approx \frac{1.3049}{0.4343} \approx 3.0009\)