Answer
$x=-\dfrac{80}{47}$
Work Step by Step
Using the properties of logarithms, the given expression, $
\log_2 x-\log_2 (3x+5)=4
,$ is equivalent to
\begin{array}{l}\require{cancel}
\log_2 \dfrac{x}{3x+5}=4
.\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}
\dfrac{x}{3x+5}=2^4
\\\\
\dfrac{x}{3x+5}=16
\\\\
x=16(3x+5)
\\\\
x=48x+80
\\\\
x-48x=80
\\\\
-47x=80
\\\\
x=\dfrac{80}{-47}
\\\\
x=-\dfrac{80}{47}
.\end{array}
Upon checking, $
x=-\dfrac{80}{47}
$ satisfies the original equation.
