回覆 5# tsusy 的帖子
\(\displaystyle \frac{1}{\ln 10}=\frac{1}{\frac{\log 10}{\log e}}=\log e\)
\(\displaystyle \approx \log 2.718 = \log \frac{2718}{1000}\)
\(\displaystyle =\log \frac{2 \times 3^2 \times 151}{10^3} \approx \log \frac{2 \times 3^2 \times 150}{10^3}\)
\(\displaystyle =\log \frac{2 \times 3^2 \times 3 \times 5 \times 10}{10^3} = \log \frac{3^3}{10}\)
\(=3 \log 3-1 \approx 3 \times 0.4771-1=0.4313\)
\(\displaystyle\frac{50}{\ln 10} \approx 50 \times 0.4313 = 21.565\)
\(\displaystyle f\left(\frac{50}{\ln 10} \right)=50 \cdot \log_{10} \left( \frac{50}{\ln 10} \right) - \frac{50}{\ln 10}\)
\(\approx 50 \cdot \log 21.565-21.565\)
\(\approx 50 \cdot \log 21.6-21.565\)
\(\displaystyle=50 \cdot \log \frac{216}{10}-21.565\)
\(\displaystyle=50 \cdot \log \frac{2^3 \times 3^3}{10}-21.565\)
\(= 50 \cdot (3 \log 2 + 3 \log 3 - 1) - 21.565\)
\(= 50 \cdot (3 \times 0.3010 + 3 \times 0.4771 - 1) - 21.565\)
\(= 45.15\)
這題用這個方法估計能算到45這個答案！
