Математика №9, №19, логарифмы

9)
$log_{ \frac{1}{2} }(1- \frac{x}{2} )+log_2 \sqrt{2- \frac{x}{4} } =0$
ОДЗ:
$\left \{ {{1- \frac{x}{2}\ \textgreater \ 0} \atop {2- \frac{x}{4} \geq 0}} \right. \\\\ \left \{ {{x\ \textless \ 2} \atop {x \leq 8}} \right. \\\\ x\ \textless \ 2 \\\\ (-\infty;2)$

$log_2 \sqrt{2- \frac{x}{4} } - log_{2}(1- \frac{x}{2} )=log_2 1 \\\\ \frac{ \sqrt{2- \frac{x}{4} }}{1- \frac{x}{2}} =1 \\\\ \sqrt{2- \frac{x}{4}} =1- \frac{x}{2} \\\\ 2- \frac{x}{4} =(1- \frac{x}{2})^2 \\\\ 2- \frac{x}{4} =1-x+ \frac{x^2}{4} \\\\ \frac{x^2}{4}- \frac{3x}{4}-1=0 \\\\ x^2-3x-4=0 \\\\ (x-4)(x+1) = 0 \\\\ x = -1$

19)
$-log_x 64-6log_8 \sqrt{x} = 7$
ОДЗ:
$x \geq 0 \\\\ x \neq 1 \\\\ x \in [0;1) \cup (1;+\infty)$

$- \frac{1}{log_{64} x} -3log_8 x= 7 \\\\ \frac{1}{log_{64} x} +3log_8 x= -7 \\\\ \frac{1}{ \frac{1}{2} log_{8} x} +3log_8 x= -7 \\\\ \frac{2}{ log_{8} x} +3log_8 x= -7 \\\\ 2 +3log^2_8 x= -7 log_{8} x \\\\ (log_8 x \neq 0) \\\\ log_8 x =a \\\\ 3a^2+7a+2 =0 \\\\ (a+2)(3a+1)=0 \\\\ log_8 x = -2; log_8 x=log_8 \frac{1}{64} ; x= \frac{1}{64} \\\\ log_8 x= - \frac{1}{3}; log_8 x=log_ 8 \frac{1}{2}; x=\frac{1}{2}$