Cho x>0 thỏa mãn {log _2}({log _8}x) = {log _8}({log _2}x)

Đặt $t = {\log _2}x,$ ta có ${\log _8}x = {\log _{{2^3}}}x = \frac{1}{3}.{\log _2}x = \frac{t}{3}$ 

suy ra ${\log _2}\frac{t}{3} = {\log _8}t \Leftrightarrow {\log _2}\frac{t}{3} = \frac{1}{3}{\log _2}t$

${\log _2}\frac{t}{3} = {\log _2}\sqrt[3]{t} \Leftrightarrow \frac{t}{3} = \sqrt[3]{t} \Leftrightarrow t = 3\sqrt 3 \Rightarrow {({\log _2}x)^2} = {t^2} = 27.$