Logarithm where $0<a<\frac{1}{2}$. Find $x$

Given that ${\mathrm{log}}_{a}(3x-4a)+{\mathrm{log}}_{a}(3x)=\frac{2}{{\mathrm{log}}_{2}a}+{\mathrm{log}}_{a}(1-2a)$ where $0<a<\frac{1}{2}$. find the value of $x$

I got the attempt until $x=\frac{2(a+\sqrt{(a-1{)}^{2}}}{3}$ and $-\frac{2(\sqrt{(a-1{)}^{2}}-a)}{3}$.

How to proceed to find $x$?

Given that ${\mathrm{log}}_{a}(3x-4a)+{\mathrm{log}}_{a}(3x)=\frac{2}{{\mathrm{log}}_{2}a}+{\mathrm{log}}_{a}(1-2a)$ where $0<a<\frac{1}{2}$. find the value of $x$

I got the attempt until $x=\frac{2(a+\sqrt{(a-1{)}^{2}}}{3}$ and $-\frac{2(\sqrt{(a-1{)}^{2}}-a)}{3}$.

How to proceed to find $x$?