#### Answer

$x=3 $

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
First, use the properties of logarithms to simplify the given logarithmic expression, $
\log_5(x+2)+\log_5(x-2)=1
,$ to a single logarithm. Then convert to exponential form. Next, use concepts of quadratic equations to solve for the variable. Finally, it is a must to do checking at the end.
$\bf{\text{Solution Details:}}$
Using $\log_b (xy)=\log_b x+\log_b y$ or the Product Rule of logarithms, the given equation is equivalent to \begin{array}{l}\require{cancel}
\log_5[(x+2)(x-2)]=1.\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}
(x+2)(x-2)=5^1 \\\\ (x+2)(x-2)=5 .\end{array}
Using the properties of equality and the concepts of quadratic equations, the solution/s to the equation above is/are \begin{array}{l}\require{cancel}
(x)^2-(2)^2=5 \\\\ x^2-4=5 \\\\ x^2=5+4 \\\\ x^2=9 \\\\ x=\pm\sqrt{9} \\\\ x=\pm3 .\end{array}
Upon checking, only $ x=3 $ satisfies the original equation.