Đáp án:
$C$
Giải thích các bước giải:
$ĐK: \left\{\begin{array}{l} \log_8x>0\\ \log_2x>0\end{array} \right.\\ \Leftrightarrow x>1\\ P=(\log_2x)^2\\ \sqrt{P}=|\log_2x|\\ \sqrt{P}=\log_2x(Do \ \log_2x>0 \forall \ x>1)\\ \log_2(\log_8x)=\log_8(\log_2x)\\ \Leftrightarrow \log_2(\log_{2^3}x)=\log_{2^3}(\log_2x)\\ \Leftrightarrow \log_2\left(\dfrac{1}{3}\log_{2}x\right)=\dfrac{1}{3}\log_{2}(\log_2x)\\ \Leftrightarrow \log_2\left(\dfrac{\log_{2}x}{3}\right)=\log_{2}(\sqrt[3]{\log_2x})\\ \Leftrightarrow \log_2\left(\dfrac{ \sqrt{P}}{3}\right)=\log_{2}\left(\sqrt[3]{ \sqrt{P}}\right)\\ \Leftrightarrow \dfrac{ \sqrt{P}}{3}=\sqrt[6]{P}\\ \Leftrightarrow \left(\dfrac{ \sqrt{P}}{3}\right)^6=\left(\sqrt[6]{P}\right)^6\\ \Leftrightarrow \dfrac{ P^3}{3^6}=P\\ \Leftrightarrow \dfrac{ P^3}{P}=3^6\\ \Leftrightarrow P^2=3^6\\ \Leftrightarrow P=3^3(Do \ P=(\log_2x)^2 >0 \forall \ x>1)\\ \Leftrightarrow P=27$
