Cho log 6 45 = a + log 2 5 + b / log 2 3 + c với a, b,c thuộc Z

Ta có \({\log _6}45 = {\log _6}\left( {{3^2}.5} \right)\)\( = \frac{{{{\log }_2}\left( {{3^2}5} \right)}}{{{{\log }_2}6}}\)\( = \frac{{2{{\log }_2}3 + {{\log }_2}5}}{{{{\log }_2}\left( {2.3} \right)}}\)\( = \frac{{2{{\log }_2}3 + {{\log }_2}5}}{{1 + {{\log }_2}3}}\)\( = \frac{{2\left( {{{\log }_2}3 + 1} \right) + {{\log }_2}5 - 2}}{{{{\log }_2}3 + 1}}\) \( = 2 + \frac{{{{\log }_2}5 - 2}}{{{{\log }_2}3 + 1}}\)

Vậy \(\left\{ \begin{array}{l}a = 2\\b =  - 2\\c = 1\end{array} \right.\) \( \Rightarrow a + b + c = 2 - 2 + 1 = 1\).