$8.29 \frac{1}{2}+\log _{9} x-\log _{3}(5 x)>\log _{\frac{1}{3}}(x+3)$. 

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$8.29 \frac{1}{2}+\log _{9} x-\log _{3}(5 x)>\log _{\frac{1}{3}}(x+3)$.
8.29 We will convert all terms to logarithms with base 9:

$$
\begin{aligned}
& \frac{1}{2}=\log _{9} 9^{\frac{1}{2}}=\log _{9} 3 ; \log _{3}(5 x)=\log _{9}\left(25 x^{2}\right) \\
& \log _{\frac{1}{3}}(x+3)=-\log _{3}(x+3)=-\log _{9}(x+3)^{2}=\log _{9}(x+3)^{-2}
\end{aligned}
$$

$$
\left\{\begin{array}{l}
x>0 \\
x+3>0 \\
\log _{9} \frac{3 x(x+3)^{2}}{25 x^{2}}>\log _{9} 1
\end{array}\right.
$$

$$
\Rightarrow\left\{\begin{array}{l}
x>0 \\
\frac{3(x+3)^{2}}{25 x}-1>0
\end{array}\right.
$$

Solving the last system, we find that $x>0$.

Answer: $(0, \infty)$.