0\\\\ \log^2_2x-4\cdot \frac{1}{2}\log_2x-3=0\\\\ \log_2^2x-2\log_2x-3=0\\ \log_2x=y;\\ \\ y^2-2y-3=0\\ \left \{ {{y_1y_2=-3} \atop {y_1+y_2=2}} \right. \left [{ {{y=3} \atop {y=-1}} \right. \left [{ {{\log_2x=3} \atop {\log_2x=-1}} \right. \left [{ {{x=8} \atop {x=\frac{1}{2}}} \right." alt="\sf \displaystyle (\log_2x)^2-4\log_4x=3; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; ODZ: x>0\\\\ \log^2_2x-4\cdot \frac{1}{2}\log_2x-3=0\\\\ \log_2^2x-2\log_2x-3=0\\ \log_2x=y;\\ \\ y^2-2y-3=0\\ \left \{ {{y_1y_2=-3} \atop {y_1+y_2=2}} \right. \left [{ {{y=3} \atop {y=-1}} \right. \left [{ {{\log_2x=3} \atop {\log_2x=-1}} \right. \left [{ {{x=8} \atop {x=\frac{1}{2}}} \right." align="absmiddle" class="latex-formula">
Ответ: 