Cho \({\log _{12}}3 = a\). Tính \({\log _{24}}18\) theo \(a\).

* Hướng dẫn giải

Ta có: \(a = {\log _{12}}3 = \frac{{{{\log }_2}3}}{{{{\log }_2}12}} = \frac{{{{\log }_2}3}}{{{{\log }_2}\left( {{2^2}.3} \right)}} = \frac{{{{\log }_2}3}}{{{{\log }_2}\left( {{2^2}} \right) + {{\log }_2}3}} = \frac{{{{\log }_2}3}}{{2 + {{\log }_2}3}} \Rightarrow {\log _2}3 = \frac{{2a}}{{1 - a}}\)

Ta có: \({\log _{24}}18 = \frac{{{{\log }_2}18}}{{{{\log }_2}24}} = \frac{{{{\log }_2}\left( {{{2.3}^2}} \right)}}{{{{\log }_2}\left( {{2^3}.3} \right)}} = \frac{{1 + 2{{\log }_2}3}}{{3 + {{\log }_2}3}} = \frac{{1 + 2.\frac{{2a}}{{1 - a}}}}{{3 + \frac{{2a}}{{1 - a}}}} = \frac{{3a + 1}}{{3 - a}}\)

 Vậy \({\log _{24}}18 = \frac{{3a + 1}}{{3 - a}}\)