# Logarithm where 0<a<1/2. Find x Given that log_a(3x-4a)+ log_a(3x)=2/(log_2a)+ log_a(1-2a) where 0<a<1/2. find the value of x.

Logarithm where $0. Find $x$
Given that ${\mathrm{log}}_{a}\left(3x-4a\right)+{\mathrm{log}}_{a}\left(3x\right)=\frac{2}{{\mathrm{log}}_{2}a}+{\mathrm{log}}_{a}\left(1-2a\right)$ where $0. find the value of $x$
I got the attempt until $x=\frac{2\left(a+\sqrt{\left(a-1{\right)}^{2}}}{3}$ and $-\frac{2\left(\sqrt{\left(a-1{\right)}^{2}}-a\right)}{3}$.
How to proceed to find $x$?
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indyterpep
So, let's use log's with base a.
${\mathrm{log}}_{a}\left(3x\left(3x-4a\right)\right)=2{\mathrm{log}}_{a}2+{\mathrm{log}}_{a}\left(1-2a\right),$
and
$3x\left(3x-4a\right)=4\left(1-2a\right).$
Solve it and
$x=\frac{2}{3},\phantom{\rule{.5em}{0ex}}x=\frac{4a-2}{3}.$
But for second $x$, $3x-4a=-2$, and log is undefined. So, you have $x=2/3$.
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