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?

Set $a = 2^x, b = 3^x$, then $6^x = (2 \cdot 3)^x = 2^x \cdot 3^x = ab$ and $2^{2x+2} = 2^2 \cdot 2^{2x} = 4 \cdot 2^x \cdot 2^x = 4a^2$, $9^{x+1} = 9 \cdot 9^x = 9 \cdot 3^x \cdot 3^x = 9b^2$. So we get $$12a^2 - 35ab + 18b^2 = 0.$$ I think.
What can you do with that?
