Điều kiện: x>0
Ta có:
\(\begin{array}{l} {\log _2}x.{\log _4}x.{\log _8}x.{\log _{16}}x = \frac{{81}}{{24}}\\ \Leftrightarrow \left( {{{\log }_2}x} \right)\left( {\frac{1}{2}{{\log }_2}x} \right)\left( {\frac{1}{3}{{\log }_2}x} \right)\left( {\frac{1}{4}{{\log }_2}x} \right) = \frac{{81}}{{24}}\\ \Leftrightarrow {\log _2}^4x = 81 \end{array}\)
Suy ra: \({t^4} = 81 \Leftrightarrow t = \pm 3\)
