#### Answer

The solution is $x=\,\,1$.

#### Work Step by Step

Consider the given equation system,
$\begin{align}
& {{\log }_{3}}x+{{\log }_{3}}\left( x+2 \right)=1 \\
& {{\log }_{3}}\left\{ x\left( x+2 \right) \right\}=1 \\
& x\left( x+2 \right)={{3}^{1}}
\end{align}$
It can be further simplified as below:
$\begin{align}
& {{x}^{2}}+2x-3=0 \\
& {{x}^{2}}+3x-x-3=0 \\
& x\left( x+3 \right)-\left( x+3 \right)=0 \\
& \left( x+3 \right)\left( x-1 \right)=0
\end{align}$
Therefore
$x=-3\,\,\text{or}\,\,1$
${{\log }_{a}}x=b$ can also be written as ${{a}^{b}}=x$, where $a>0$ so for any value of b, x is always positive. Therefore, discard the negative value of the logarithm.
The solutions of the equations are $x=\,\,1$.