Log^2 по основанию 2 (3-x) + log^3 по основанию корень из 2 (3-x) меньше либо равно 0

$\log^2_2(3-x)+\log^3_{\sqrt{2}}(3-x) \leq 0\\ \log^2_2(3-x)+8\log^3_2}(3-x) \leq 0\\ \log^2_2(3-x)(1+8\log_2(3-x)) \leq 0\\ \log_2(3-x) = t\\ t^2(1+8t) \leq 0\\ t_1=t_2=0,t_3=-\frac{1}{8}\\$

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$t \in (-\infty;-\frac{1}{8}]\\ t \leq -\frac{1}{8}\\\\ \log_2(3-x) \leq -\frac18\\\\ 3-x \leq 2^{-\frac18}\\ 3-x \leq \frac{1}{\sqrt[8]2}\\ -x \leq \frac{1}{\sqrt[8]2}-3\\ x \geq 3-\frac{1}{\sqrt[8]2}\\\\ 3-x \ \textgreater \ 0\\ -x \ \textgreater \ -3\\ x \ \textless \ 3\\\\ x \in [3-\frac{1}{\sqrt[8]2};3)\\$