Логарифмические неравенства
\[{\text{Решите систему:}}\] $%\left\{ \begin{array}{l}{2^x} \le 3\\{\log _x}\left( {\frac{3}{2}x - 1} \right) \ge - 1\end{array} \right.$%
$$\eqalign{{\log _{\sqrt {x - 1} }}\left( {31x - 61\sqrt {x - 1} - 1} \right) > 2} $$
$$\eqalign{ \frac{{x + 1}}{{3 - {{\log }_3}\left( {9 - {3^{ - x}}} \right)}} \leqslant 1 } $$
$$\eqalign{ \frac{{\left| {9 - {2^x}} \right| - 7}}{{\sqrt {{{\log }_2}\left( {\frac{x}{4} + \frac{1}{x}} \right)} }} < 0 } $$
\[2{\log _5}\left( {2x} \right) - {\log _5}\frac{x}{{1 - x}} \leqslant {\log _5}\left( {8{x^2} + \frac{1}{x} - 3} \right)\]
\[2{\log _2}\left( {x\sqrt 5 } \right) - {\log _2}\frac{x}{{1 - x}} \leqslant {\log _2}\left( {5{x^2} + \frac{1}{x} - 2} \right)\]
\[{\log _{\sqrt x }}{\left( {x - 12} \right)^2} + 1 \geqslant \log _x^2\left( {12x - {x^2}} \right)\]