Đáp án:
\[C\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\log _{{a^b}}}c = \frac{1}{b}{\log _a}c\\
{\log _{2018}}x + {\log _{\sqrt {2018} }}x + {\log _{\sqrt[3]{{2018}}}}x + .... + {\log _{\sqrt[{2018}]{{2018}}}}x = \frac{{2019}}{2}\\
\Leftrightarrow {\log _{2018}}x + {\log _{{{2018}^{\frac{1}{2}}}}}x + {\log _{{{2018}^{\frac{1}{3}}}}}x + ..... + {\log _{{{2018}^{\frac{1}{{2018}}}}}}x = \frac{{2019}}{2}\\
\Leftrightarrow {\log _{2018}}x + 2{\log _{2018}}x + 3{\log _{2018}}x + .... + 2018{\log _{2018}}x = \frac{{2019}}{2}\\
\Leftrightarrow {\log _{2018}}x.\left( {1 + 2 + 3 + ... + 2018} \right) = \frac{{2019}}{2}\\
\Leftrightarrow \frac{{2019.2018}}{2}{\log _{2018}}x = \frac{{2019}}{2}\\
\Leftrightarrow {\log _{2018}}x = \frac{1}{{2018}}\\
\Leftrightarrow x = {2018^{\frac{1}{{2018}}}}\\
\Leftrightarrow x = \sqrt[{2018}]{{2018}}
\end{array}\)
