Question

㏒²(-㏒x)+㏒㏒²x≤3 ( у всех логарифмов основание 2)

Answer 1

$log_2^2(-log_2x)+log_2, log_2^2x leq 3; ;\\ ODZ:; -log_2x textgreater 0; ,; log_2x textless 0; ,; ; ; underline {0 textless x textless 1}\\(-log_2x)^2=log_2^2x; ; ;; ; t=log_2x; ; to \\log_2^2(-t)+log_2((-t)^2)-3 leq 0\\log_2^2(-t)+2cdot log_2(-t)-3 leq 0\\z=log_2(-t); ;; ; z^2+2z-3 leq 0; ;; -3 leq z leq 1\\ left { {{log_2(-t) leq 1} atop {log_2(-t) geq -3}} right. ; left { {{-t leq 2} atop {-t geq frac{1}{8}}} right. ; left { {{t geq -2} atop {t leq -frac{1}{8}} right.$

$left { {{log_2x geq -2} atop {log_2x leq -frac{1}{8}}} right. ; left { {{x geq 2^{-2}} atop {x leq 2^{-frac{1}{8}}}} right. ; left { {{x geq frac{1}{4}} atop {x leq frac{1}{sqrt[8]{2}}}} right. \\Otvet:quad xin [, frac{1}{4}; ;; frac{1}{sqrt[8]{2}}, ]$