Nếu log _126 = a và log _127 = b thì

${\log _{12}}6 = a \Rightarrow {\log _6}12 = \frac{1}{a} \Rightarrow 1 + {\log _6}2 = \frac{1}{a} \Rightarrow {\log _6}2 = \frac{{1 - a}}{a} \Rightarrow {\log _2}6 = \frac{a}{{1 - a}}.$
$\begin{array}{l} {\log _{12}}7 = b \Rightarrow {\log _7}12 = \frac{1}{b} \Rightarrow {\log _7}2.{\log _2}6 + {\log _7}2 = \frac{1}{b} \Rightarrow {\log _7}2\left( {\frac{a}{{1 - a}} + 1} \right) = \frac{1}{b}\\ \Rightarrow {\log _7}2 = \frac{{1 - a}}{b} \Rightarrow {\log _2}7 = \frac{b}{{1 - a}}. \end{array}$