Answer
$x=\dfrac{26-8\sqrt{10}}{9}\approx0.078$
and
$x=\dfrac{26+8\sqrt{10}}{9}\approx5.700$
Work Step by Step
Recall:
(1) The Change-of-Base Formula for Logarithms: $\log_a{b}=\dfrac{\log_c{b}}{\log_c{a}}$
(2) $\log_a{a^b}=b$
(3) $n\cdot \log_a{b} = \log_a{(b^n)}$
(4) $\log_a{b} - \log_a{c} = \log_a{\left(\frac{b}{c}\right)}$
Use the change-of-base formula above to obtain:
\begin{align*}
\log_2{(3x+2)}-\dfrac{\log_2{x}}{\log_2{4}}&=3\\
\log_2{(3x+2)}-\dfrac{\log_2{x}}{\log_2{(2^2)}}&=3\\
\end{align*}
Use the rule in (2) above to obtain:
\begin{align*}
\log_2{(3x+2)}-\dfrac{\log_2{x}}{2}&=3\\
2\left(\log_2{(3x+2)}-\dfrac{\log_2{x}}{2}\right)&=3(2)\\
2\log_2{(3x+2)}-\log_2{x}&=6\\
\end{align*}
Use the rules in (3) and (4) above to obtain:
\begin{align*}
\log_2{(3x+2)^2}-\log_2{x}&=6\\
\log_2{\left(\frac{(3x+2)^2}{x}\right)}&=6
\end{align*}
Use the rule $\log_a{b}=y \longrightarrow a^y=b$ to obtain:
\begin{align*}
\frac{(3x+2)^2}{x}&=2^6\\
\frac{(3x)^2+2(3x)(2)+2^2}{x}&=64\\
\frac{9x^2+12x+4}{x}&=64\\
x\left(\frac{9x^2+12x+4}{x}\right)&=64(x)\\
9x^2+12x+4&=64x\\
9x^2+12x+4-64x&=0\\
9x^2-52x+4&=0
\end{align*}
Solve the quadratic equation using the quadratic formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $a=9, b=-52, \text{ and } c=4$ to obtain:
\begin{align*}
x&=\frac{-(-52)\pm\sqrt{(-52)^2-4(9)(4)}}{2(9)}\\
&=\frac{52\pm\sqrt{2704-144}}{18}\\
&=\frac{52\pm\sqrt{2560}}{18}\\
&=\frac{52\pm\sqrt{256(10)}}{18}\\
&=\frac{52\pm\sqrt{16^2(10)}}{18}\\
&=\frac{52\pm16\sqrt{10}}{18}\\
&=\frac{26\pm8\sqrt{10}}{9}
\end{align*}
Therefore, the solutions are:
$x=\dfrac{26-8\sqrt{10}}{9}\approx0.078$
and
$x=\dfrac{26+8\sqrt{10}}{9}\approx5.700$