Answer
$x=\dfrac{2}{3}$
Work Step by Step
Using the properties of logarithms, the given expression, $
\log_2 x+\log_2 (3x+1)=1
,$ is equivalent to
\begin{array}{l}\require{cancel}
\log_2 [x(3x+1)]=1
\\\\
\log_2 (3x^2+x)=1
.\end{array}
Since $y=b^x$ is equivalent to $\log_b y=x$, then the solution to the equation, $
\log_4 \dfrac{x}{2x-3}=3
,$ is
\begin{array}{l}\require{cancel}
3x^2+x=2^1
\\\\
3x^2+x=2
\\\\
3x^2+x-2=0
\\\\
(3x-2)(x+1)=0
\\\\
x=\left\{ -1,\dfrac{2}{3} \right\}
.\end{array}
Upon checking, only $
x=\dfrac{2}{3}
$ satisfies the original equation.
