If the approximate value of $ \log _{ 10 }{ (4.04) }$is $0. \ abcdef,$ it is given that $\log _{ 10 }{ (4) }=0.6021$ &$ \log _{ 10 }{ (e) }=0.4343,$ then the value of abcd must be

Using the linear approximation for logarithms, $\log(a + h) \approx \log(a) + \frac{h}{a \ln(10)}$, we can set $a = 4$ and $h = 0.04$. Since $\log_{10}(e) = \frac{1}{\ln(10)} \approx 0.4343$, we have $\log_{10}(4.04) \approx 0.6021 + \frac{0.04}{4} \times 0.4343$. This evaluates to $0.6021 + 0.01 \times 0.4343 = 0.6021 + 0.004343 = 0.606443$, meaning the first four digits after the decimal are $6064$.

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