## Precalculus (6th Edition) Blitzer

The solution is $\left\{ 2 \right\}$.
Solve the provided equation as follows: \begin{align} & \log \left( x+3 \right)+\log x=1 \\ & log\left[ \left( x+3 \right)\left( 3 \right) \right]=1 \\ & \left( x+3 \right)x={{10}^{1}} \\ & {{x}^{2}}+3x=10 \end{align} Further solve this equation \begin{align} & {{x}^{2}}+3x-10=0 \\ & {{x}^{2}}+5x-2x-10=0 \\ & x\left( x+5 \right)-2\left( x+5 \right)=0 \\ & \left( x+5 \right)\left( x-2 \right)=0 \end{align} Finally, we get, $x=-5,2$ Put the value of $x=-5$ in the original equation, $\log \left( -5+3 \right)+\log \left( -5 \right)=1$ Since, log of a negative number is not defined, $x=-5$ is not solution Thus the solution set is $\left\{ 2 \right\}$.