#### Answer

$x=\dfrac{7}{2}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\log_3(x+1)-\log_3(x-3)=2
,$ use the properties of logarithms to simplify the expression at the left. Then convert to exponential form and use the properties of equality to isolate the variable.
$\bf{\text{Solution Details:}}$
Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
\log_3 \dfrac{x+1}{x-3}=2
.\end{array}
Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{x+1}{x-3}=3^2
\\\\
\dfrac{x+1}{x-3}=9
.\end{array}
Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{x+1}{x-3}=\dfrac{9}{1}
\\\\
(x+1)(1)=(x-3)(9)
\\\\
x+1=9x-27
\\\\
x-9x=-27-1
\\\\
-8x=-28
\\\\
x=\dfrac{-28}{-8}
\\\\
x=\dfrac{7}{2}
.\end{array}