The approximate value of log _{10} 998 is (gi | vedclass

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(D) Let $f(x) = \log_{10} x = \frac{\log_e x}{\log_e 10} = (\log_{10} e)(\log_e x) = 0.4343(\log_e x)$.On differentiating with respect to $x$,we get $

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$2.9991314$

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(D) Let $f(x) = \log_{10} x = \frac{\log_e x}{\log_e 10} = (\log_{10} e)(\log_e x) = 0.4343(\log_e x)$.On differentiating with respect to $x$,we get $f'(x) = \frac{0.4343}{x}$.Let $x = 998 = 1000 - 2 = a + h$.Here,$a = 1000$ and $h = -2$.$f(a) = f(1000) = \log_{10}(1000) = 3 \log_{10} 10 = 3$.Also,$f'(a) = f'(1000) = \frac{0.4343}{1000} = 0.0004343$.Using the approximation formula $f(a+h) \approx f(a) + hf'(a)$:$\log_{10}(998) \approx 3 + (-2)(0.0004343) = 3 - 0.0008686 = 2.9991314$.

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

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