# How do you solve for x?: 2log_3(x) = 3 log_3(4)

Oct 22, 2015

I found $x = 8$

#### Explanation:

We can use a rule of the log to write:
${\log}_{3} \left({x}^{\textcolor{red}{2}}\right) = {\log}_{3} \left({4}^{\textcolor{red}{3}}\right)$

then take $3$ to the power of the right and left side to cancel the logs:
${\textcolor{b l u e}{3}}^{{\log}_{3} \left({x}^{\textcolor{red}{2}}\right)} = {\textcolor{b l u e}{3}}^{{\log}_{3} \left({4}^{\textcolor{red}{3}}\right)}$
${\cancel{\textcolor{b l u e}{3}}}^{\cancel{{\log}_{3}} \left({x}^{\textcolor{red}{2}}\right)} = {\cancel{\textcolor{b l u e}{3}}}^{\cancel{{\log}_{3}} \left({4}^{\textcolor{red}{3}}\right)}$
and get:
${x}^{2} = {4}^{3} = 64$
$x = \pm \sqrt{64} = \pm 8$
we can use only the positive one $x = 8$.