Answer
$x=2$ and $x=-4$
Work Step by Step
Recall the definition of a logarithm (pg. 451):
$\log_{b}{x}=y$ iff $b^y=x$
Applying this definition, we get:
$2^3=x^2+2x$
$8=x^2+2x$
$0=x^2+2x-8$
$0=(x-2)(x+4)$
Use the Zero Product Property by equating each factor to zero, then solve each equation to obtain:
\begin{align*}
x-2&=0 &\text{or}& &x+4=0\\
x&=2 &\text{or}& &x=-4
\end{align*}
We check the answers:
For $x=2$:
$\log_2 (2^2+2*2)=3$
$\log_2 (4+4)=3$
$\log_2 (8)=3$
$\log_2 (2^3)=3$
$3=3$
For $x=-4$:
$\log_2 ((-4)^2+2*-4)=3$
$\log_2 (16-8)=3$
$\log_2 (8)=3$
$\log_2 (2^3)=3$
$3=3$
Thus, the solutions are $x=2$ and $x=-4$.
