# How do you solve log_3 (2x-4) - log_3 (x-1)=log_3 (x-2)?

Dec 21, 2015

I found:
$x = 3$

#### Explanation:

We can use the property of logs:
$\log x - \log y = \log \left(\frac{x}{y}\right)$
And write:
${\log}_{3} \left[\frac{2 x - 4}{x - 1}\right] = {\log}_{3} \left(x - 2\right)$
If the logs are equal the arguments must be as well, so:
$\frac{2 x - 4}{x - 1} = x - 2$
$2 x - 4 = \left(x - 1\right) \left(x - 2\right)$
$2 x - 4 = {x}^{2} - 2 x - x + 2$
${x}^{2} - 5 x + 6 = 0$
${x}_{1 , 2} = \frac{5 \pm \sqrt{25 - 24}}{2} =$
${x}_{1} = 3$ YES
×_2=2 NO