<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] The following substitution may have to work:
$$2^x=t; ~~3^x=s$$
Note that the equation simplifies to, $$12 t^2-35st+ 18s^2=0$$
This factorises to $$(3t-2s)(4t-9s)=0$$
Therefore,
$$3\cdot2^x=2 \cdot 3^x ~~\text{or}~~2^{x+2}=3^{x+2}$$ Since, $(2,3)=1$, we have that $\boxed{x=1~~ \text{or}~~-2}$
Edited to add:
You can view that as a quadratic equation in $t$:
So, the solution will have to be, $$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24}$$
This is a bit numerically taxing and honestly, I did not do it this way. Rather, I resorted to something that is equivalent to this. You need to write $18 \times 12$ as product of two numbers whose sum is $35$. Later, prefix a minus sign to these numbers and use them here.
With a little bit of playing around, with factorisations, you'll see they should be $27 \times 8$. 

Added by dindoun
$$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24}
=\dfrac{35s\pm\sqrt{s^2[(-35)^2-4(12)(18)]}}{24}
=\frac{35\pm 54}{24}s = \frac{2}{3}s\ or \frac{9}{4}s$$
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