Log1/3(x^2 +8x)=-2Помогите решить!

ОДЗ
$x^{2} +8x\ \textgreater \ 0 \\ \\ x(x+8)\ \textgreater \ 0$

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__________-8_________0______________

x∈(-∞; -8)∪(0; +∞)

$log_{ \frac{1}{3}} (x^2 +8x)=-2 \\ \\ x^{2} +8x=( \frac{1}{3})^{-2} \\ \\ x^{2} +8x=9 \\ \\ x^{2} +8x-9=0 \\ \\ x^{2} -x+9x-9=0 \\ \\ x(x-1)+9(x-1)=0 \\ \\ (x+9)(x-1)=0 \\ \\ x_1=-9 \\ \\ x_2=1$

Ответ 9; -1