Use $ \log_{10} 2 \equal{} 0.301,\ \log_{10} 3 \equal{} 0.477,\ \log_{10} 7 \equal{} 0.845$ to find the value of $ \log_{10} (10!)$.
Note that you must answer according to the rules:fractional part of $ 0.5$ and higher is rounded up, and everything strictly less than $ 0.5$ is rounded down,
say $ 1.234\longrightarrow 1.23$. Then find the minimum integer value $ n$ such that $ 10! < 2^{n}$.