Precalculus (6th Edition) Blitzer

Solution set = $\{5\}$
For the equation to be defined, logarithms must have positive arguments $\begin{array}{ll} x+4 \gt 0, & \\ x \gt -4 & (*) \end{array}$ $(*)$ is the condition eventual solutions must satisfy. $2\log_{3}(x+4)=\log_{3}9+2\qquad$... LHS: power rule, RHS: $\log_{a}a^{n}=n$ $\log_{3}(x+4)^{2}=\log_{3}9+\log_{3}3^{2}\qquad$ ... RHS: product rule $\log_{3}(x+4)^{2}=\log_{3}(9\cdot 3^{2})\qquad$ ... if $\log_{b}M=\log_{b}N,$ then M=N ... logarithmic functions are one to one, $(x+4)^{2}=81$ $x+4=\pm 9$ $x=-4\pm 9$ $x=-13\qquad$ ... does not satisfy (*) - not a solution $x=5\qquad$ ... satisfies (*) Solution set = $\{5\}$