Ответ:
Объяснение:
0} \atop {4/x\neq1 }} \right. \\ \\ x\in(0;4)U(4;+oo)\\ \\log_{2}x-log_{2} 8-1=\frac{1}{log_{2}4-log_{2}x} \\ \\ log_{2}x-4=\frac{1}{2-log_{2}x } \\ \\(log_{2}x-4)(log_{2}x-2)=-1\\ \\(log_{2}x )^2-6log_{2}x+9=0\\ \\ (log_{2}x-3)^2=0\\ \\log_{2}x=3\\ \\x=2^3=8\\ \\ otvet:x=8" alt="\left \{ {{x>0} \atop {4/x\neq1 }} \right. \\ \\ x\in(0;4)U(4;+oo)\\ \\log_{2}x-log_{2} 8-1=\frac{1}{log_{2}4-log_{2}x} \\ \\ log_{2}x-4=\frac{1}{2-log_{2}x } \\ \\(log_{2}x-4)(log_{2}x-2)=-1\\ \\(log_{2}x )^2-6log_{2}x+9=0\\ \\ (log_{2}x-3)^2=0\\ \\log_{2}x=3\\ \\x=2^3=8\\ \\ otvet:x=8" align="absmiddle" class="latex-formula">