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?
 A: 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
$$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24}
=\dfrac{35s\pm\sqrt{s^2[(-35)^2-4(12)(18)]}}{24}
=\frac{35\pm 54}{24}s = \frac{2}{3}s\ or \frac{9}{4}s$$
A: $$
\begin{eqnarray}
0 &=& 3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1}
\\&=& 12 \cdot (2^x)^2 - 35 \cdot (2^x3^x) + 18 \cdot (3^x)^2
\\&=& 12 t^2 - 35 st + 18 s^2
\qquad\text{for}\qquad s=3^x,~t=2^x
\\&=& (3t - 2s)(4t - 9s)
\\
\implies&&
  s=\frac32t \quad\text{or}\quad
  s=\frac49t
\\&& s=\left(\frac23\right)^\delta t
  \quad\text{for}
  \quad\delta=\frac{1\pm3}{2}=-1~\text{or}~2
\\&& 3^x=\left(\frac23\right)^\delta 2^x
\\&& 1=\left(\frac23\right)^{x+\delta}
\\&& 0=(x+\delta)\log\left(\frac23\right)
\\&& x+\delta=0
\\&& x=-\delta=-\frac{1\pm3}{2}=\boxed{~1~\text{or}~-2~}
\end{eqnarray}
$$
To make the factorization in the fourth line,
from the signs of the coefficients $12,-35,18$,
we can see that we need to find
a combination of the factors
from a row on the left
and a row on the right
$$
\begin{eqnarray}
   1 \cdot 12 &\quad& 1 \cdot 18
\\ 2 \cdot  6 &\quad& 2 \cdot  9
\\ 3 \cdot  4 &\quad& 3 \cdot  6
\end{eqnarray}
$$
so that a sum of the products
(either of the first numbers and the second numbers,
or of the first numbers with the second numbers)
is $35$. So taking $3 \cdot 4$ for $12$ and $2 \cdot 9$ for $18$,
we can obtain $2\cdot 4+3\cdot 9=8+27=35$ to get the factors above.
A: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0 \Rightarrow 12 \cdot 2^{2x} - 35 \cdot 2^x \cdot 3^x + 18 \cdot 3^{2x} = 0 \Rightarrow$
$\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 .
A: 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?
A: from user21436 you have 
$$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24}
=\dfrac{35s\pm\sqrt{s^2[(-35)^2-4(12)(18)]}}{24}
=\frac{35\pm 54}{24}s = \frac{2}{3}s\ or \frac{9}{4}s
$$ 
