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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?

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\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*}

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  • $\begingroup$ How do you go from row nr2 to row nr3?, the last factoring there? $\endgroup$ – F88 Apr 13 '16 at 19:47
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    $\begingroup$ Replace $3^x$ by $a$ and $2^x$ by $b$, you have $a^2-ab-2b^2=(a+b)(a-2b)$. $\endgroup$ – Ng Chung Tak Apr 13 '16 at 19:49
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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.

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