Đáp án:
$S =\left\{10^{\displaystyle{\dfrac{\log_610+2}{3}}}\right\}$
Giải thích các bước giải:
$\quad \log(\log6) + \log(\log x^3 - 2)= 0\qquad \left(ĐK: x > \sqrt[3]{100}\right)$
$\Leftrightarrow \log[\log6(\log x^3 -2)]= 0$
$\Leftrightarrow \log6(\log x^3 - 2)= 1$
$\Leftrightarrow \log x^3 - 2 = \log_610$
$\Leftrightarrow 3\log x = \log_610 + 2$
$\Leftrightarrow \log x = \dfrac{\log_610+2}{3}$
$\Leftrightarrow x = 10^{\displaystyle{\dfrac{\log_610+2}{3}}}$ (nhận)
Vậy $S =\left\{10^{\displaystyle{\dfrac{\log_610+2}{3}}}\right\}$
