If log _{10} 87.5 = 1.9421,then the number of | vedclass

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(A) Let $x = (875)^{10}$.We can write $875 = 87.5 	imes 10$.So,$x = (87.5 	imes 10)^{10}$.Taking $\log_{10}$ on both sides:$\log_{10} x = \log_{10}

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$30$

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(A) Let $x = (875)^{10}$.We can write $875 = 87.5 	imes 10$.So,$x = (87.5 	imes 10)^{10}$.Taking $\log_{10}$ on both sides:$\log_{10} x = \log_{10} (87.5 	imes 10)^{10} = 10 	imes \log_{10} (87.5 	imes 10)$.Using the property $\log(ab) = \log a + \log b$:$\log_{10} x = 10 	imes (\log_{10} 87.5 + \log_{10} 10)$.Given $\log_{10} 87.5 = 1.9421$ and $\log_{10} 10 = 1$:$\log_{10} x = 10 	imes (1.9421 + 1) = 10 	imes 2.9421 = 29.421$.The characteristic of $\log_{10} x$ is $29$. The number of digits in $x$ is given by $	ext{characteristic} + 1$.Number of digits $= 29 + 1 = 30$.

If $\log _{10} 87.5 = 1.9421$,then the number of digits in $(875)^{10}$ is

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