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.
On
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 $$x = \log_3 a = \frac{\ln 3}{\ln 3/2} \log_3 2 = \frac{\ln 2}{\ln 3/2} = \frac{\ln 2}{\ln 3 - \ln 2}$$
On
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: $$3^x=2^{x+1}\iff 2=x\ln 3=(x+1)\ln2\iff x=\frac{\ln2}{\ln 3-\ln 2}.$$
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