Cho \({\log _{12}}27 = a\). Tính \(T = {\log _{36}}24\) theo \(a\).

* Hướng dẫn giải

Ta có \({\log _{12}}27 = \frac{3}{{{{\log }_3}\left( {{2^2}.3} \right)}} = a\)

Suy ra \(\frac{3}{{1 + 2{{\log }_3}2}} = a\) hay \({\log _3}2 = \frac{{3 - a}}{{2a}}\) ( \(a \ne 0\) vì \(a = {\log _{12}}27 > {\log _{12}}1\) ).

Khi đó \({\log _{36}}24 = \frac{{{{\log }_3}24}}{{{{\log }_3}36}} = \frac{{3{{\log }_3}2 + 1}}{{2{{\log }_3}2 + 2}} = \frac{{1 + \frac{{9 - 3a}}{{2a}}}}{{2 + \frac{{6 - 2a}}{{2a}}}} = \frac{{9 - a}}{{6 + 2a}}\)