## College Algebra (11th Edition)

$x=3$
$\bf{\text{Solution Outline:}}$ 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.