Answer
$x=1$
Work Step by Step
Recall:
(1) $\log_a{x} + \log_a{y}=\log_a{xy}$
(2) $\log_a{x} = \log_a{y} \longrightarrow x=y$
Use rule (1) above to obtain:
\begin{align*}
\log_a{[x(x-2)]}&=\log_a{(x+4)}\\
\log_a{(x^2-2x)}&=\log_a{(x+4)}\\
\end{align*}
Use rule (2) above to obtain:
\begin{align*}
x^2-2x&=x+4\\
x^2-2x-x-4&=0\\
x^2-3x-4&=0\\
(x-4)(x+1)&=0
\end{align*}
Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain:
\begin{align*}
x-4&=0 &\text{or}& &x+1=0\\
x&=4 &\text{or}& &x=-1
\end{align*}
Since in $\log_a{x}$, $x\gt0$, then $x=-1$ is an extraneous solution.
Thus, the solution is $x=4$.
