\(\log_52=\frac{\log_32}{\log_35}=\frac{1}{\log_23}:\frac{1}{\log_53}=\frac{\log_53}{\log_23}=\frac{b}{a}\)
Ta có: \(\log_645=\log_{\left(2\cdot3\right)}\left(3^2\cdot5\right)=\log_{2\cdot3}3^2+\log_{2\cdot3}5=2\cdot\log_{2\cdot3}3+\frac{1}{\log_53\cdot2}\)
\(=2\cdot\frac{1}{\log_32\cdot3}+\frac{1}{\log_53+\log_52}\)
\(=2\cdot\frac{1}{\log_32+\log_33}+\frac{1}{\log_53+\log_52}=2\cdot\frac{1}{\log_32+1}+\frac{1}{\log_53+\log_52}\)
\(=\frac{2}{\frac{1}{\log_23}+1}+\frac{1}{b+\frac{b}{a}}=\frac{2}{\frac{1}{a}+1}+\frac{1}{b+\frac{b}{a}}\)
\(=\frac{2}{\frac{1}{a}+\frac{a}{a}}+\frac{1}{\frac{ba+b}{a}}=2:\frac{a+1}{a}+1:\frac{ba+b}{a}\)
\(=\frac{2a}{a+1}+\frac{a}{ba+b}=\frac{2a}{a+1}+\frac{a}{b\left(a+1\right)}=\frac{2ab+a}{b\left(a+1\right)}=\frac{a\left(2b+1\right)}{b\left(a+1\right)}\)
=>m=1; n=2; p=1
=>m+n+p=1+2+1=4
