Cho a,b là các số dương thỏa mãn log9 a = log16 b

Đặt \({\log _9}a = {\log _{16}}b = {\log _{12}}\frac{{5b - a}}{2} = t\)

\( \Rightarrow \left\{ \begin{array}{l}a = {9^t}\\b = {16^t}\\\frac{{5b - a}}{2} = {12^t}\end{array} \right.\)

\( \Rightarrow {5.16^t} - {9^t} = {2.12^t}\)

\( \Leftrightarrow {\left( {\frac{3}{4}} \right)^{2t}} + 2.{\left( {\frac{3}{4}} \right)^t} - 5 = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}{\left( {\frac{3}{4}} \right)^t} = - 1 + \sqrt 6 \\{\left( {\frac{3}{4}} \right)^t} = - 1 - \sqrt 6 \left( {KTM} \right)\end{array} \right.\).

Do đó \(\frac{a}{b} = {\left( {\frac{3}{4}} \right)^{2t}} = {\left( { - 1 + \sqrt 6 } \right)^2} = 7 - 2\sqrt 6 \).