What is x if log_3 (2x-1) = 2 + log_3 (x-4)?

Nov 26, 2015

$x = 5$

Explanation:

We will use the following:

• ${\log}_{a} \left(b\right) - {\log}_{a} \left(c\right) = {\log}_{a} \left(\frac{b}{c}\right)$
• ${a}^{{\log}_{a} \left(b\right)} = b$

${\log}_{3} \left(2 x - 1\right) = 2 + {\log}_{3} \left(x - 4\right)$

$\implies {\log}_{3} \left(2 x - 1\right) - {\log}_{3} \left(x - 4\right) = 2$

$\implies {\log}_{3} \left(\frac{2 x - 1}{x - 4}\right) = 2$

$\implies {3}^{{\log}_{3} \left(\frac{2 x - 1}{x - 4}\right)} = {3}^{2}$

$\implies \frac{2 x - 1}{x - 4} = 9$

$\implies 2 x - 1 = 9 x - 36$

$\implies - 7 x = - 35$

$\implies x = 5$

Nov 26, 2015

I found: $x = 5$

Explanation:

We can start writing it as:
${\log}_{3} \left(2 x - 1\right) - {\log}_{3} \left(x - 4\right) = 2$
use the property of the logs: $\log x - \log y = \log \left(\frac{x}{y}\right)$ and write:
${\log}_{3} \left(\frac{2 x - 1}{x - 4}\right) = 2$
use the definition of log:
${\log}_{b} x = a \to x = {b}^{a}$
to get:
$\frac{2 x - 1}{x - 4} = {3}^{2}$ rearranging:
$2 x - 1 = 9 \left(x - 4\right)$
$2 x - 9 x = - 36 + 1$
$7 x = 35$
$x = \frac{35}{7} = 5$