## Precalculus (6th Edition) Blitzer

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