## College Algebra (11th Edition)

$x=\left\{ -8,0 \right\}$
Changing to exponential form, the given expression, $\log_2[(2x+8)(x+4)]=5 ,$ is equivalent to \begin{array}{l}\require{cancel} (2x+8)(x+4)=2^5 .\end{array} Using concepts of quadratic equations, the solution/s to the equation above is/are \begin{array}{l}\require{cancel} 2x(x)+2x(4)+8(x)+8(4)=32 \\\\ 2x^2+8x+8x+32-32=0 \\\\ 2x^2+16x=0 \\\\ \dfrac{2x^2+16x}{2}=\dfrac{0}{2} \\\\ x^2+8x=0 \\\\ x(x+8)=0 .\end{array} Equating each factor to zero and then solving for the variable, the solutions are $x=\left\{ -8,0 \right\} .$ Upon checking, $x=\left\{ -8,0 \right\}$ satisfy the original equation.