Question

РОЗВЯЗАТИ НЕРІВНІСТЬ ІЗ LOG!!!1
ДАМ 60 БАЛОВ!!!

Answer 1

$\displaystyle\bf\\ODZ:\\3-x>0\\\\-x>-3\\\\x<3 \ \ \Rightarrow \ \ x\in(-\infty \ ; \ 3)\\\\\\\log_{2} ^{2} (3-x)+\log_{\sqrt[3]{2} }(3-x)\leq 0\\\\\log_{2} ^{2} (3-x)+\log_{2} (3-x)^{3} \leq 0\\\\\log_{2} ^{2} (3-x)+3\log_{2} (3-x)\leq 0\\\\\log_{2} (3-x)=m\\\\m^{2} +3m\leq 0\\\\m(m+3)\leq 0$

+ + + + + [- 3] - - - - - [0] + + + + +    m

              //////////////////

$\displaystyle\bf\\\left \{ {{m\geq -3} \atop {m\leq 0}} \right. \\\\\\\left \{ {{\log_{2} (3-x)\geq -3} \atop {\log_{2}(3-x)\leq 0 }} \right.\\\\\\\left \{ {{3-x\geq \frac{1}{8} } \atop {3-x\leq1 }} \right. \\\\\\\left \{ {{-x\geq -2\frac{7}{8} } \atop {-x\leq -2}} \right. \\\\\\\left \{ {{x\leq 2\frac{7}{8} } \atop {x\geq 2}} \right.\\\\\\Otvet:\Big[2 \ ; \ 2\frac{7}{8} \Big]$