Logarithmic inequality online

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$$- \log{\left(16 - 2 x \right)} + \log{\left(x + 27 \right)} > \log{\left(x \right)}$$

$$\left(0 < x \wedge x < 3\right) \vee \left(\frac{9}{2} < x \wedge x < 8\right)$$ $$x\ in\ \left(0, 3\right) \cup \left(\frac{9}{2}, 8\right)$$