Nếu \({\log _2}3 = a\) thì \({\log _{27}}108\) bằng

* Hướng dẫn giải

Ta có: \({\log _{72}}108 = {\log _{72}}\left( {36.3} \right) = {\log _{72}}36 + {\log _{72}}3 = \frac{1}{{{{\log }_{36}}72}} + \frac{1}{{{{\log }_3}72}}\) 

+) \({\log _{36}}72 = {\log _{36}}\left( {36.2} \right) = {\log _{36}}36 + {\log _{{6^2}}}2 = 1 + \frac{1}{2}{\log _6}2\) 

\( = 1 + \frac{1}{2}.\frac{1}{{{{\log }_2}6}} = 1 + \frac{1}{2}.\frac{1}{{{{\log }_2}2 + {{\log }_2}3}} = 1 + \frac{1}{2}.\frac{1}{{1 + a}} = \frac{{3 + 2a}}{{2 + 2a}}\) 

\( + ){\log _3}72 = {\log _3}\left( {{2^3}{{.3}^2}} \right) = 3{\log _3}2 + 2{\log _3}3 = \frac{3}{a} + 2 = \frac{{3 + 2a}}{a}\) 

Suy ra \({\log _{72}}108 = \frac{{2 + 2a}}{{3 + 2a}} + \frac{a}{{3 + 2a}} = \frac{{2 + 3a}}{{3 + 2a}}\)