Cho số thực $\Large x$ thỏa mãn $\Large \mathrm{log}_2(\mathrm{log}_8x

Cho số thực $\Large x$ thỏa mãn $\Large \mathrm{log}_2(\mathrm{log}_8x)=\mathrm{log}_8(\mathrm{log}_2x)$. Tính giá trị $\Large P=(\mathrm{log}_2x)^4$.

• A.

  $\Large P=27$.

• B.

  $\Large P=81\sqrt{3}$.

• C.

  $\Large P=729$.

• D.

  $\Large P=243$.

Chọn C
Điều kiện: $\Large \left\{\begin{align} & x > 0 \\ & \mathrm{log}_8x > 0 \\ & \mathrm{log}_2x > 0 \end{align}\right.$

Ta có: $\Large \mathrm{log}_2(\mathrm{log}_8x)=\mathrm{log}_8(\mathrm{log}_2x)$ $\Large \Leftrightarrow \mathrm{log}_2\left(\dfrac{1}{3}\mathrm{log}_2x\right)=\mathrm{log}_2(\mathrm{log}_2x)^{\frac{1}{3}}$

$\Large \Leftrightarrow \dfrac{1}{3}\mathrm{log}_2x=\sqrt[3]{\mathrm{log}_2x}$ $\Large \Leftrightarrow \mathrm{log}_2x=3\sqrt[3]{\mathrm{log}_2x}$ (*).

Đặt $\Large t=\sqrt[3]{\mathrm{log}_2x}$ $\Large (t > 0)$ $\Large \Rightarrow t^3=\mathrm{log}_2x$

(*) $\Large \Leftrightarrow t^3=3t$ $\Large \Leftrightarrow \left[\begin{align} & t=0 \\ & t=\sqrt{3} \\ & t=-\sqrt{3} \end{align}\right.$ $\Large \Rightarrow t=\sqrt{3}$ $\Large \Rightarrow \sqrt[3]{\mathrm{log}_2x}=\sqrt{3}$ $\Large \Leftrightarrow \mathrm{log}_2x=3\sqrt{3}$

$\Large \Leftrightarrow x=2^{3\sqrt{3}}$ (thỏa mãn đề bài).

Vậy $\Large P=(3\sqrt{3})^4=729$.