The approximate value of log _{10} 1002 is (G | vedclass

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(B) We use the linear approximation formula: $f(a+h) \approx f(a) + h f'(a)$.Let $f(x) = \log_{10} x$.Then $f'(x) = \frac{1}{x \ln 10} = \frac{\log_{1

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$3.0009$

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(B) We use the linear approximation formula: $f(a+h) \approx f(a) + h f'(a)$.Let $f(x) = \log_{10} x$.Then $f'(x) = \frac{1}{x \ln 10} = \frac{\log_{10} e}{x}$.Given $a = 1000$ and $h = 2$,we have:$f(1002) \approx f(1000) + 2 f'(1000)$.$f(1000) = \log_{10} 1000 = 3$.$f'(1000) = \frac{\log_{10} e}{1000} = \frac{0.4343}{1000} = 0.0004343$.Therefore,$\log_{10} 1002 \approx 3 + 2(0.0004343)$.$\log_{10} 1002 \approx 3 + 0.0008686 = 3.0008686$.Rounding to four decimal places,we get $3.0009$.

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

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