Tình giá trị biểu thức S=a+2b+3c biết {log _7}12 = x; log _{12}}24 = y; log _{54}}168=(axy+1)/(bxy+cx)

\(\begin{array}{l} \bullet \,\,\left\{ \begin{array}{l}{\log _7}12 = x\\{\log _{12}}24 = y\end{array} \right. \Rightarrow xy = {\log _7}12.{\log _{12}}24 = {\log _7}24\\ \Rightarrow {\log _{54}}168 = \frac{{{{\log }_7}168}}{{{{\log }_7}54}} = \frac{{{{\log }_7}\left( {24.7} \right)}}{{{{\log }_7}54}} = \frac{{{{\log }_7}24 + {{\log }_7}7}}{{{{\log }_7}54}} = \frac{{xy + 1}}{{{{\log }_7}54}} \Rightarrow a = 1.\\ \bullet \,\,b{\rm{x}}y + c{\rm{x}} = {\log _7}54 \Leftrightarrow b{\log _7}24 + c{\log _7}12 = {\log _7}54 \Leftrightarrow {\log _7}\left( {{{24}^b}{{.12}^c}} \right) = {\log _7}54\\ \Leftrightarrow {24^b}{.12^c} = 54 \Leftrightarrow c = {\log _{12}}\frac{{54}}{{{{24}^b}}} \Rightarrow \left\{ \begin{array}{l}b =  - 5\\c = 8\end{array} \right. \Rightarrow P = a + 2b + 3c = 1 + 2.\left( { - 5} \right) + 3.8 = 15.\end{array}\)