Nếu \({\log _{12}}6 = a;{\rm{ }}{\log _{12}}7 = b\) thì :

* Hướng dẫn giải

Ta có \({\log _2}7 = \frac{{{{\log }_{12}}7}}{{{{\log }_{12}}2}} = \frac{{{{\log }_{12}}7}}{{{{\log }_{12}}12 - {{\log }_{12}}6}} = \frac{b}{{1 - a}}\)