2018 AMC 12B Problem #7

$\begin{aligned} P & = \log_3 7\log_5 9 \log_7 11 \log_9 13 \cdots \log_{21} 25 \log_{23} 27 \\ & = \frac {\log 7 \log 9 \log 11 \log 13 \cdots \log 25 \log 27} {\log 3 \log 5 \log 7 \log 9 \cdots \log 21 \log 23} \\ & = \frac {\cancel{\log 7} \cancel{\log 9} \cancel{\log 11} \cancel{\log 13} \cdots \log 25 \log 27}{\log 3 \log 5 \cancel{\log 7} \cancel{\log 9} \cdots \cancel{\log 21} \cancel{\log 23}} \\ & = \frac {\log 25 \log 27}{\log 3 \log 5} \\ & = \frac {2 \cancel{\log 5} \times 3 \cancel{\log 3}}{\cancel{\log 3} \cancel{\log 5}} \\ & = \boxed{6} \end{aligned}$

$\log_3 7 \cdot \log_5 9 \cdot \log_7 11 \cdot \log_9 13 \cdots \log_{21} 25 \cdot \log_{23} 27=$

$\frac{ln 7}{ln 3}\cdot\frac{ln 9}{ln 5}\cdot\frac{ln 11}{ln 7}\cdot\frac{ln 13}{ln 9}\cdots\frac{ln 25}{ln 21}\cdot\frac{ln 27}{ln 23}=$

$\frac1{ln 3}\cdot\frac1{ln 5}\cdot\frac{ln 25}1\cdot\frac{ln 27}1=\frac{ln 27}{ln 3}\cdot\frac{ln 25}{ln 5}=\frac{3\cdot ln 3}{ln 3}\cdot\frac{2\cdot ln 5}{ln 5}=3\cdot2=\boxed6$