An algebra problem by Naren Bhandari

Divide through by $9^{x}$ to get the equation

$\dfrac{2^{2x}}{3^{2x}} + \dfrac{2^{x}3^{x}}{3^{x}3^{x}} = 1 \Longrightarrow \left(\dfrac{2}{3}\right)^{2x} + \left(\dfrac{2}{3}\right)^{x} = 1 \Longrightarrow y^{2} + y - 1 = 0$ where $y = \left(\dfrac{2}{3}\right)^{x}$ .

Then as clearly $y \gt 0$ we take the positive root $y = \dfrac{-1 + \sqrt{5}}{2} = \phi - 1$ , and so

$\ln(\phi - 1) = x\ln\left(\dfrac{2}{3}\right) \Longrightarrow x = \dfrac{\ln(\phi - 1)}{\ln(2) - \ln(3)} = \boxed{1.187}$ to 3 decimal places.