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

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