## College Algebra (11th Edition)

$x=\left\{ 0,6 \right\}$
Changing to exponential form, the given equation, $\log_2(x-4)(x-2)=3 ,$ is equivalent to \begin{array}{l}\require{cancel} (x-4)(x-2)=2^3 \\\\ (x-4)(x-2)=8 .\end{array} Using the concepts of solving quadratic equations, then \begin{array}{l}\require{cancel} x(x)+x(-2)-4(x)-4(-2)=8 \\\\ x^2-2x-4x+8=8 \\\\ x^2-2x-4x=8-8 \\\\ x^2-6x=0 \\\\ x(x-6)=0 .\end{array} Equating each factor to $0$ and solving for the variable, the solutions are $x=\left\{ 0,6 \right\} .$ Upon checking, $x=\left\{ 0,6 \right\}$ satisfy the original equation.