# How do you solve log_3x-2log_3 2=3log_3 3 ?

Feb 29, 2016

I found : $x = 108$

#### Explanation:

We can use the properties of logs that tell us:
$\log x - \log y = \log \left(\frac{x}{y}\right)$
and also:
$\log {x}^{a} = a \log x$
in our case we get:
${\log}_{3} x - {\log}_{3} \left({2}^{2}\right) = {\log}_{3} \left({3}^{3}\right)$
then:
${\log}_{3} \left(\frac{x}{2} ^ 2\right) = {\log}_{3} \left(27\right)$
if the logs must be equal the arguments must be as well. So, we can write:
$\frac{x}{4} = 27$
$x = 4 \cdot 27 = 108$