## Precalculus (6th Edition) Blitzer

Solution set = $\{2\}.$
For the equation to be defined, logarithms must have positive arguments $\left[\begin{array}{llll} x \gt 0, & and & x+3 \gt 0 & \\ & & x\gt-3 & \end{array}\right]$ $x \gt -3 \qquad (*)$ This is the condition eventual solutions must satisfy. $\log x+\log(x+3)=\log 10\qquad$... LHS: product rule $\log(x^{2}+3x)=\log 10 \quad$ ... $\log$ is one-to-one $x^{2}+3x=10$ $x^{2}+3x-10=0$ Factor: +5 and -2 are factors of -10 whose sum is +3. $(x+5)(x-2)=0$ $x=2 \qquad$ ... satisfies (*) $x=-5 \quad$ ... does not satisfy (*) Thus, the solution set = $\{2\}.$