#### Answer

$x=8 $

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
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.