## College Algebra (11th Edition)

$x=8$
$\bf{\text{Solution Outline:}}$ First, use the properties of logarithms to simplify the given logarithmic expression, $\log_2(x-7)+\log_2 x=3 ,$ 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 expression is equivalent to \begin{array}{l}\require{cancel} \log_2[(x-7)x]=3 .\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-7)x=2^3 \\\\ (x-7)x=8 .\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^2-7x=8 \\\\ x^2-7x-8=0 \\\\ (x-8)(x+1)=0 .\end{array} Equating each factor to $0$ and then solving for the variable, then the solutions are $x=\left\{ -1,8 \right\} .$ Upon checking, only $x=8$ satisfies the original equation.