The approximate value of $\log _{10} 998$ is (given that $\log _{10} \mathrm{e}=0.4343$ )

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Aicharya · Accepted Answer

Let $\begin{aligned} \mathrm{f}(x)=\log _{10} x=\frac{\log _{\mathrm{e}} x}{\log _{\mathrm{e}} 10} & =\left(\log _{10} \mathrm{e}ight)\left(\log _{\mathrm{e}} xight) \ & =0.4343\left(\log _{\mathrm{e}} xight) \end{aligned}$ On differentiating w.r.t. $x$, we get $\begin{aligned} \mathrm{f}^{\prime}(x) & =\frac{0.4343}{x} \ 	ext { Let } x & =998 \ & =1000-2=\mathrm{a}+\mathrm{h} \ 	herefore \quad \mathrm{a}= & 1000, \mathrm{~h}=-2 \ \mathrm{f}(\mathrm{a}) & =\mathrm{f}(1000) \ & =\log _{10}(1000) \ & =3 \log _{10} 10 \ 	herefore \quad \mathrm{f}(\mathrm{a}) & =3 \end{aligned}$ $\begin{aligned} & 	ext { Also, } \mathrm{f}^{\prime}(\mathrm{a})=\mathrm{f}^{\prime}(1000)=\frac{0.4343}{1000}=0.0004343 \ & \mathrm{f}(\mathrm{a}+\mathrm{h}) \approx \mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a}) \ & 	herefore \quad \log _{10}(998) \approx 3-2(0.0004343) \ & \end{aligned}$

The approximate value of $\log _{10} 998$ is (given that $\log _{10} \mathrm{e}=0.4343$ )

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