СРОЧНО!!!!!!!!!!!!!!!!!!!!!!!!!!!!

2. $log_{ \sqrt{2} } (32 \sqrt{2}) =log_{ \sqrt{2} } 32+ log_{ \sqrt{2} } (\sqrt{2})= log_{ 2^{ \frac{1}{2}} } (2^{5} ) + 1= 2 log_{ 2}(2^{5} ) + 1= \\ = 2*5+1 = 11 \\ \sqrt{13^{log_{13}81} } + 12^{log_{ \sqrt{12} }10} = \sqrt{81 } +12^{2log_{ 12}10}= \\ =9+ (12^{log_{ 12}10})^{2} = 9+10^{2} = 109 \\$

3. 0} \atop {3-x=x+2}} \right. \\ \left \{ {{x<3} \atop {x= \frac{1}{2} }} \right. \\ x= 0,5 \\ " alt="lg(3-x)=lg4(x+2) \\ \\ \left \{ {{3-x>0} \atop {3-x=x+2}} \right. \\ \left \{ {{x<3} \atop {x= \frac{1}{2} }} \right. \\ x= 0,5 \\ " align="absmiddle" class="latex-formula">
$log_{3^{2} } x = 4log_{3} x \\ \frac{1}{2} log_{3 } x = 4log_{3} x \\ log_{3 } x^{ \frac{1}{2} } = log_{3}x^{4} \\ x^{ \frac{1}{2} } = x^{4} \\ x=1$

4, $log_{0,3}(3x-1) \geq log_{0,3}(3+x) \\$
Данное неравенство равносильно системе:
0}} \right. \\ \left \{ {{3x-x \leq 3+1} \atop {3x>1}} \right. \\ \left \{ {{2x \leq 4} \atop {x> \frac{1}{3} }} \right. \\ \left \{ {{x \leq 2} \atop {x> \frac{1}{3} }} \right. \\" alt=" \left \{ {{3x-1 \leq 3+x} \atop {3x-1>0}} \right. \\ \left \{ {{3x-x \leq 3+1} \atop {3x>1}} \right. \\ \left \{ {{2x \leq 4} \atop {x> \frac{1}{3} }} \right. \\ \left \{ {{x \leq 2} \atop {x> \frac{1}{3} }} \right. \\" align="absmiddle" class="latex-formula">
Ответ: ( 1/3 ; 2 ]