Giải bài 67 trang 124 SGK giải tích nâng cao 12

Điều kiện $x>0$

$\begin{aligned} a)\,& {{\log }_{2}}x+{{\log }_{4}}x={{\log }_{\frac{1}{2}}}\sqrt{3} \\ & \Leftrightarrow {{\log }_{2}}x+\dfrac{1}{2}{{\log }_{2}}x+\dfrac{1}{2}{{\log }_{2}}x=0 \\ & \Leftrightarrow 2{{\log }_{2}}x=0 \\ & \Leftrightarrow x=1 \\ b)\, & {{\log }_{\sqrt{3}}}x.{{\log }_{3}}x.{{\log }_{9}}x=8 \\ & \Leftrightarrow 2.{{\log }_{3}}x.{{\log }_{3}}x.\dfrac{1}{2}.{{\log }_{3}}x=8 \\ & \Leftrightarrow {{\left( {{\log }_{3}}x \right)}^{3}}={{2}^{3}} \\ & \Leftrightarrow {{\log }_{3}}x=2 \\ & \Leftrightarrow x={{3}^{2}}=9 \\ \end{aligned}$