Answer
The solution is $\left\{ 3 \right\}$.
Work Step by Step
$\begin{align}
& {{\log }_{3}}\left[ x\left( x+6 \right) \right]=3 \\
& x\left( x+6 \right)=\text{anti}{{\log }_{3}}\left( 3 \right) \\
& x\left( x+6 \right)={{3}^{3}}
\end{align}$
Solving further, we get,
$\begin{align}
& {{x}^{2}}+6x=27 \\
& {{x}^{2}}+6x-27=0 \\
& {{x}^{2}}+9x-3x-27=0 \\
& \left( x+9 \right)\left( x-3 \right)=0
\end{align}$
Therefore, $x=-9,3$
Put $x=-9$ in the original equation.
${{\log }_{3}}\left( -9 \right)+{{\log }_{3}}\left( -9+6 \right)=3$
The log of a negative number is not defined. Thus, $x=-9$ is an extraneous solution.
Thus, $\left\{ 3 \right\}$ is the only solution.
