## College Algebra (6th Edition)

Domain: $(-\infty,-6)\cup(-6,\infty)$.
Domain restriction for logarithmic functions: the argument of the logarithm must be positive. $h(x)=\log_{3}(x+6)^{2}$ $(x+6)^{2}>0$ A square of a real number is nonnegative, so the only problem is when the LHS is zero: $x+6=0$ $x=-6\qquad$ ... $-6$ is the only number excluded Domain: $(-\infty,-6)\cup(-6,\infty)$.