Nếu $a > 0,b > 0$ thỏa mãn ${\log _4}a = {\log _6}b = {\log _9}\left( {a + b} \right)$ thì $\dfrac{a}{b}$ bằng:

Ta có: ${\log _4}a = {\log _6}b = {\log _9}(a + b) = t$ suy ra $\left\{ \begin{array}{l}a = {4^t}\\b = {6^t}\\a + b = {9^t}\end{array} \right.$

$\Rightarrow {4^t} + {6^t} = {9^t}$$\Leftrightarrow {\left( {\dfrac{2}{3}} \right)^{2t}} + {\left( {\dfrac{2}{3}} \right)^t} - 1 = 0$  

Đặt ${\left( {\dfrac{2}{3}} \right)^t} = u > 0 \Rightarrow {u^2} + u - 1 = 0$ $\Rightarrow \left[ \begin{array}{l}u = \dfrac{{ - 1 + \sqrt 5 }}{2}\left( {tm} \right)\\u = \dfrac{{ - 1 - \sqrt 5 }}{2}\left( {ktm} \right)\end{array} \right.$

Nên ${\left( {\dfrac{2}{3}} \right)^t} = \dfrac{{ - 1 + \sqrt 5 }}{2}$

Mà $\dfrac{a}{b} = \dfrac{{{4^t}}}{{{6^t}}} = {\left( {\dfrac{2}{3}} \right)^t}$  nên $\dfrac{a}{b} = \dfrac{{ - 1 + \sqrt 5 }}{2}$