## Precalculus (6th Edition) Blitzer

$x=2$
Consider the given expression: ${{\log }_{2}}x+{{\log }_{2}}\left( x+2 \right)=3$ Using the product identity, we get ${{\log }_{a}}A+{{\log }_{b}}B={{\log }_{ab}}\left( AB \right)$ So, \begin{align} & {{\log }_{2}}x+{{\log }_{2}}\left( x+2 \right)=3 \\ & \log \left( x\left( x+2 \right) \right)=3 \end{align} Use anti log. Then, \begin{align} & {{\log }_{2}}\left( x\left( x+2 \right) \right)=3 \\ & x\left( x+2 \right)={{2}^{3}} \\ & {{x}^{2}}+2x=8 \\ & {{x}^{2}}+2x-8=0 \end{align} This implies that. \begin{align} & {{x}^{2}}-2x+4x-8=0 \\ & x(x-2)+4(x-2)=0 \\ & \left( x-2 \right)\left( x+4 \right)=0 \\ & x=2,x=-4 \end{align} We eliminate the $-4$ solution, because we can not take the log of a negative number.