## Precalculus (6th Edition) Blitzer

Solution set = $\{-3\}$
For the equation to be defined, logarithms must have positive arguments $\left[\begin{array}{lll} x+6 \gt 0, & and & x+4 \gt 0\\ x \gt -6 & & x \gt -4 \end{array}\right]$ $x \gt -4 \qquad (*)$ This is the condition eventual solutions must satisfy. $\log_{3}(x+6)+\log_{3}(x+4)=2$ LHS: apply the product rule; RHS write $2$ as $\log_{3}3^{2}$ $\log_{5}[(x+6)(x+4)]=\log_{3}3^{2}\qquad$ (if $\log_{b}M=\log_{b}N,$ then M=N) $(x+6)(x+4)=3^{2}$ $x^{2}+10x+24=9$ $x^{2}+10x+15=0$ Factor, by finding two factors of $15$ whose sum is $10$ $(x+5)(x+3)=0$ $x=-5$ or $x=-3$ The solution $x=-5$ does not satisfy (*), so the solution set = $\{-3\}$