$\log_a(3x-4a)+\log_a(3x)=\frac2{\log_2a}+\log_a(1-2a)$; $0
Given that
  $$\log_a(3x-4a)+\log_a(3x)=\frac2{\log_2a}+\log_a(1-2a)$$
  where $0<a<\frac12$, find $x$.
My question is how do we find the value of $x$ but we don't know the exact value of $a$?  
 A: Hint. Recall the main properties of the logarithm.
Then we have that $\frac{1}{\log_2a}=\log_a 2$. 
Moreover for $x>4a/3>0$ (the argument of the logarithm should be positive),
$$\log_a((3x-4a)\cdot 3x))=\log_a(2^2\cdot(1-2a))$$
which is equivalent (logarithm is an injective function) to 
$$(3x-4a)\cdot 3x=2^2\cdot(1-2a)$$
or the quadratic equation
$$9x^2 - 12ax - 4(1-2a)=0.$$
Can you take it from here?
P.S. We expect to find two solutions. At end remember to check the condition $x>4a/3$!
A: $$\log_a(3x-4a)+\log_a(3x)=\frac2{\log_2a}+\log_a(1-2a)$$
$$\log_a(3x(3x-4a))=\frac2{\log_aa/\log_a2}+\log_a(1-2a)$$
$$\log_a(3x(3x-4a))=2\log_a2+\log_a(1-2a)$$
$$\log_a(3x(3x-4a))=\log_a4(1-2a)$$
$$3x(3x-4a)=4(1-2a)$$
$$9x^2-12ax+8a-4=0$$
$$x=\frac{12a\pm\sqrt{144a^2-4\cdot9(8a-4)}}{18}$$
$$=\frac{12a\pm12(a-1)}{18}=\frac{2a\pm2(a-1)}3=\frac{4a-2}3\lor\frac23$$
A: Put $u=3x-2a$ to exploit the symmetry and note that $\frac 1{\log_2}a=\log_a 2$:
$$\begin{align}
\log_a(3x-4a)+\log_a(3x)&=\frac2{\log_2a}+\log_a(1-2a)\\
\log_a (\underbrace{3x-4a}_{u-2a})(\underbrace{3x}_{u+2a})&=\log_a 2^2(1-2a)\\
(u-2a)(u+2a)&=4(1-2a)\\
u^2-4a^2&=4(1-2a)\\
u^2&=4(a-1)^2\\
u=3x-2a&=\pm 2(a-1)\\
x&=\frac {4a-2}3 \text{ or }\frac 23 \\
\end{align}$$
Note that 
$$0<a<\frac 12\\
0<4a<2\\
-2<4a-2<0\\
-\frac 23<\frac {4a-2}3<0$$
Hence $$\color{red}{-\frac 23<x<0, \;\; x=\frac 12}$$
A: $x$ might come out as a function of $a$.
$$(3x-4a)(3x) = 4(1-2a) \implies 9x^2 - 12ax - 4(1-2a) = 0$$
Solve this quadratic equation.
