What is the solution to the equation $9^x - 6^x - 2\cdot 4^x = 0 $?

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

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 $$x = \log_3 a = \frac{\ln 3}{\ln 3/2} \log_3 2 = \frac{\ln 2}{\ln 3/2} = \frac{\ln 2}{\ln 3 - \ln 2}$$

Hint: Can you factorise the quadratic?