# 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?

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*}
• Replace $3^x$ by $a$ and $2^x$ by $b$, you have $a^2-ab-2b^2=(a+b)(a-2b)$. – Ng Chung Tak Apr 13 '16 at 19:49
dividing by $$2^{2x}$$ we get $$\left(\frac{3^x}{2^x}\right)^2-\frac{3^x}{2^x}-2=0$$ and set $$t=\frac{3^x}{2^x}$$ and you will get an quadratic equation.