<s>[INST] Solving the exponential equation: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ I have this exponential equation that I don't know how to solve:
$3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ with $x \in \mathbb{R}$
I tried to factor out a term, but it does not help. Also, I noticed that:
$2 \cdot 9^{x+1} = 2 \cdot 3^{2x+2}$
and tried to write the polynomial as a binomial square, without success.
I know I should solve it using logarithm, but I don't see how to continue.  
EDIT: WolframAlpha factors it as: $(3 \cdot 2^x - 2 \cdot 3^x)(2^{x+2} - 3^{x+2}) = 0$ and then the solution is straightforward. Any hint about how to reach that?
 [/INST] $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0 \Rightarrow 12 \cdot 2^{2x} - 35 \cdot 2^x \cdot 3^x + 18 \cdot 3^{2x} = 0 \Rightarrow$
$\Rightarrow 12 \cdot \left(\frac{2}{3}\right)^{2x}-35 \cdot \left(\frac{2}{3}\right)^{x}+18=0 $
Now make substitution : $\left(\frac{2}{3}\right)^{x} = t$ , and solve quadratic equation .
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