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

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(C) Let $x = 3^{12} \times 2^8$. Taking $\log_{10}$ on both sides: $\log_{10} x = \log_{10} (3^{12} \times 2^8)$ $= 12 \log_{10} 3 + 8 \log_{10} 2$ $= 12(0.47712) + 8(0.30103)$ $= 5.72544 + 2.40824$ $= 8.13368$. The number of digits in $x$ is given by $\lfloor \log_{10} x \rfloor + 1$. $= \lfloor 8.13368 \rfloor + 1$ $= 8 + 1 = 9$.