Cho hai số thực dương \(a,b\) thỏa mãn \({{\log }_{4}}a={{\log }_{6}}b={{\log }_{9}}\left( a+b \right).\) Tính \(\frac{a}{b}.\)

* Hướng dẫn giải

Đặt \({{\log }_{4}}a={{\log }_{6}}b={{\log }_{9}}\left( a+b \right)=t.\)

\( \Leftrightarrow \left\{ \begin{array}{l} {\log _4}a = t\\ {\log _6}b = t\\ {\log _9}\left( {a + b} \right) = t \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = {4^t}\\ b = {6^t}\\ a + b = {9^t} \end{array} \right..\)

Ta có \({4^t} + {6^t} = {9^t} \Leftrightarrow {\left( {\frac{4}{9}} \right)^t} + {\left( {\frac{2}{3}} \right)^t} - 1 = 0 \Leftrightarrow \left[ \begin{array}{l} {\left( {\frac{2}{3}} \right)^t} = \frac{{ - 1 + \sqrt 5 }}{2}\\ {\left( {\frac{2}{3}} \right)^t} = \frac{{ - 1 - \sqrt 5 }}{2}\left( {VN} \right) \end{array} \right. \Leftrightarrow \frac{a}{b} = \frac{{ - 1 + \sqrt 5 }}{2}.\)