Question

Срочно $log _{2} frac{8}{x} - frac{10}{log _{2}16x } geq 0$

Answer 1

$log_{2} (frac{8}{x})-frac{10}{log_{2}(16x)}geq0 \ \ left { {{frac{8}{x} textgreater 0} atop {16x textgreater 0}} right. = textgreater x textgreater 0 \\ log_{2} 8- log_{2} x-frac{10}{log_{2}(16)+log_{2}x} geq0 \ \ 3-log_{2}x-frac{10}{4+log_{2}x}geq0 \ \ [log_{2}x=t = textgreater 3-t-frac{10}{t+4}geq0 \ \ frac{3*(t+4)-t*(t+4)-10}{t+4}geq0 \ \ frac{3t+12-t^2-4t-10}{t+4}geq0 \ \ frac{-t^2-t+2}{t+4}geq0 | : (-1) \ \ frac{t^2+t-2}{t+4}leq 0 \ frac{(t-1)(t+2)}{t+4} leq 0$
<смотрите приложенный скриншот>
$left { {{t textless -4} atop {-2 leq t leq 1}} right. = textgreater left { {{log_{2} x textless -4} atop {-2 leq log_{2} x leq 1}} right. = textgreater left { {{{x textless 2^{-4}} atop {2^{-2} leq x leq 2^1}} right. = textgreater \ \ left { {{x textless frac{1}{16}} atop {frac{1}{4} leq x leq 2}} right. \ \ x textgreater 0 = textgreater x in (0; frac{1}{16}) cup [frac{1}{4}; 2].$