Solving exponential equation (quadratic type) I fail trying to solve the following equation:
$9^x-6^x-2^{2x+1}=0$

Trying to write it as a quadratic equation makes my constant term exponential
$(3^x)^2-2^x3^x-2^{2x+1}=0$

How can I solve this type of problem most efficiently?

\begin{align*}
  9^x-6^x-2^{2x+1} &=0 \\
  (\color{red}{3^x})^2-(\color{red}{3^x})(\color{blue}{2^x})-
  2(\color{blue}{2^x})^2 &=0 \\
  (\color{red}{3^x}+\color{blue}{2^x})
  (\color{red}{3^x}-2\cdot \color{blue}{2^x}) &= 0
\end{align*}
Rejecting $\color{red}{3^x}+\color{blue}{2^x}=0$, we have
\begin{align*}
  \left( \frac{3}{2} \right)^{x} &= 2 \\
  x(\ln 3-\ln 2) &= \ln 2 \\
  x &= \frac{\ln 2}{\ln 3-\ln 2}
\end{align*}
