Cho \({\log _6}45 = a + \frac{{{{\log }_2}5 + b}}{{{{\log }_2}3 + c}}\), với \(a,\,b,\,c \in \mathbb{Z}\). Tính tổng \(a + b + c\).

Ta có: \[\left\{ \begin{array}{l}{\log _3}a + {\log _4}{b^2} = 5\\{\log _3}{a^3} - {\log _4}b = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\log _3}a + 2{\log _4}b = 5\\3{\log _3}a - {\log _4}b = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\log _3}a = 1\\{\log _4}b = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 3\\b = 16\end{array} \right.\]

\[ \Rightarrow H = a + b = 3 + 16 = 19\].

\({\log _x}w = 12\)\( \Rightarrow {\log _w}x = \frac{1}{{12}}\)

\({\log _y}w = 20\)\( \Rightarrow {\log _w}y = \frac{1}{{20}}\).

Lại do

\({\log _{xyz}}w = 6\)\( \Leftrightarrow \frac{1}{{{{\log }_{_w}}\left( {xyz} \right)}} = 6\)\( \Leftrightarrow \frac{1}{{{{\log }_{_w}}x + {{\log }_{_w}}y + {{\log }_{_w}}z}} = 6\)\( \Leftrightarrow \frac{1}{{{{\log }_{_w}}x + {{\log }_{_w}}y + {{\log }_{_w}}z}} = 6\)

\( \Leftrightarrow \frac{1}{{\frac{1}{{12}} + \frac{1}{{20}} + {{\log }_{_w}}z}} = 6\)\( \Leftrightarrow {\log _{_w}}z = \frac{1}{{30}}\)\( \Rightarrow {\log _z}w = 30\).