If log_{10} 2 = 0.30103 and log_{10} 3 = 0.47 | vedclass

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(C) Let $y = 3^{12} 	imes 2^8$. Taking $\log_{10}$ on both sides: $\log_{10} y = \log_{10} (3^{12} 	imes 2^8) = 12 \log_{10} 3 + 8 \log_{10} 2$. Sub

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$9$

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(C) Let $y = 3^{12} 	imes 2^8$. Taking $\log_{10}$ on both sides: $\log_{10} y = \log_{10} (3^{12} 	imes 2^8) = 12 \log_{10} 3 + 8 \log_{10} 2$. Substituting the given values: $\log_{10} y = 12(0.47712) + 8(0.30103)$. $\log_{10} y = 5.72544 + 2.40824 = 8.13368$. The number of digits in a number $y$ is given by $\lfloor \log_{10} y floor + 1$. Number of digits $= \lfloor 8.13368 floor + 1 = 8 + 1 = 9$.

If $\log_{10} 2 = 0.30103$ and $\log_{10} 3 = 0.47712$,the number of digits in $3^{12} \times 2^8$ is

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