What is the solution to the equation $9^x - 6^x - 2\cdot 4^x = 0 $? I want to solve: $$9^x - 6^x - 2\cdot 4^x = 0 $$
I was able to get to the equation below by substituting $a$ for $3^x$ and $b$ for $2^x$:
$$ a^2 - ab - 2b^2 = 0 $$
And then I tried 
\begin{align}x &= \log_3{a} \\
x &= \log_2{b} \\
\log_3{a} &=\log_2{b}\end{align}
But I don't know what to do after this point. Any help is appreciated.
 A: The more elegant way to solve this equation was given by lab bhattacharjee, but I'll simply elaborate on how to solve it using your method, which was still correct. 
Hint: We have that $$a^2 - ab -2b^2 = 0 \iff b = \frac{a}{2}, \text{or }b= -a.$$

You now have a system of two equations, $\log_3 a = \log_2 b$ and from this you can see that neither $a$ or $b$ can be negative. So you need to rule out $b=-a$ as a solution. Instead, we take $b= a/2$. Substituting this into the logarithmic equation gives $$\log_3 a = \log_2 \frac{a}{2} \implies \large a = 2^{\frac{\ln 3}{\ln 3/2}}$$
Then $$x = \log_3 a = \frac{\ln 3}{\ln 3/2} \log_3 2 = \frac{\ln 2}{\ln 3/2} = \frac{\ln 2}{\ln 3 - \ln 2}$$
A: Hint: Can you factorise the quadratic?
A: HINT:
Divide by $4^x$  to get $$a^2-a-2=0$$  where $a=\left(\dfrac32\right)^x$
Can you solve for $a?$
Now for real $x,a>0$
See also : Exponent Combination Laws
A: It is a homogeneous quadratic polynomial in two variables. The first thing to do is to de-homogenise it:
$$a^2-ab-2b^2=b^2\biggl(\frac{a^2}{b^2}-\frac ab-2\biggr)$$
So setting $t=\dfrac ab$, you have to solve $\;t^2-t-2=0$, which has $-1$ as a root, hence the other is $2$, en the relation between $a$ and $b$ is:
$$a=-b\quad\text{or}\quad a=2b$$
As $x$ is a real number, the first solution can't happen. So we have:
$$3^x=2^{x+1}\iff 2=x\ln 3=(x+1)\ln2\iff x=\frac{\ln2}{\ln 3-\ln 2}.$$
