The approximate value of log _{10} 99 is (Giv | vedclass

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(C) Let $f(x) = \log _{10} x = \frac{\log _{e} x}{\log _{e} 10}$.Then,$f'(x) = \frac{1}{x \log_{e} 10} = \frac{1}{x} \log_{10} e$.Let $a = 100$ and $h

Vedclass · Accepted Answer

$1.9957$

Vedclass · Answer

(C) Let $f(x) = \log _{10} x = \frac{\log _{e} x}{\log _{e} 10}$.Then,$f'(x) = \frac{1}{x \log_{e} 10} = \frac{1}{x} \log_{10} e$.Let $a = 100$ and $h = -1$,so that $a + h = 99$.We know the approximation formula $f(a + h) \approx f(a) + h f'(a)$.Here,$f(a) = \log_{10} 100 = 2$.$f'(a) = \frac{1}{100} \log_{10} e = \frac{0.4343}{100} = 0.004343$.Substituting these values into the formula:$f(99) \approx 2 + (-1)(0.004343) = 2 - 0.004343 = 1.995657$.Rounding to four decimal places,we get $1.9957$.

The approximate value of $\log _{10} 99$ is (Given $\log _{10} e = 0.4343$)

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