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

Question

Given that $\log_a(3x-4a)+\log_a(3x)=\frac{2}{\log_2a}+\log_a(1-2a)$ 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 $x$?

Answer 1

So, let's use log's with base $a$. $$\log_a \big(3x(3x-4a)\big)=2\log_a 2 + \log_a(1-2a),$$ and $$3x(3x-4a)=4(1-2a).$$ Solve it and $$x=\frac23,\enspace x=\frac{4a-2}{3}.$$ But for second $x$, $3x-4a=-2$, and log is undefined. So, you have $x=2/3$.

Answer 2

$\log_a(3x-4a)+\log_a(3x)=\frac{2}{\log_2a}+\log_a(1-2a)$

$\log_a \left[ \frac{(3x-4a)(3x)}{1-2a} \right] = \frac{2}{\log_2a}$

$\log_a \left( \frac{9x^2-12a}{1-2a}\right) = \frac{2}{\log_2a} = 2 \times \frac{\log_a2}{\log_aa}$

$\log_a\left( \frac{9x^2-12a}{1-2a} \right) = 2 \log_a2 = \log_a4$

$\frac{9x^2-12a}{1-2a} = a^{\log_a4}=4$

$9x^2-12a = 4-8a \Rightarrow a=\frac{9x^2-4}{4}$

$0<a<\frac{1}{2} \Rightarrow 0 <9x^2-4<2 \Rightarrow \frac{2}{3}<x<\sqrt{\frac{2}{3}}$