Tìm số nguyên dương \(n\) thỏa mãn\({{\log }_{a}}2...

B

- Hướng dẫn giải

Phương pháp giải:

Sử dụng các công thức biến đổi loga

\({{\log }_{{{a}^{n}}}}b=\frac{1}{n}{{\log }_{a}}b,{{\log }_{a}}\left( b.c \right)={{\log }_{a}}b+{{\log }_{a}}c\).

Giải chi tiết:

Ta có : \(VT={{\log }_{a}}2017+\frac{1}{{{2}^{2}}}{{\log }_{\sqrt{a}}}2017+\frac{1}{{{2}^{4}}}{{\log }_{\sqrt[4]{a}}}2017+\frac{1}{{{2}^{6}}}{{\log }_{\sqrt[8]{a}}}2017+\ldots +\frac{1}{{{2}^{2n}}}{{\log }_{\sqrt[{{2}^{n}}]{a}}}2017\)

\(={{\log }_{a}}2017+\frac{1}{2}{{\log }_{a}}2017+\frac{1}{{{2}^{2}}}{{\log }_{a}}2017+\frac{1}{{{2}^{3}}}{{\log }_{a}}2017+\ldots +\frac{1}{{{2}^{n}}}{{\log }_{a}}2017\)

\(=\left( 1+\frac{1}{2}+\frac{1}{{{2}^{2}}}+\ldots +\frac{1}{{{2}^{n}}} \right).{{\log }_{a}}2017=\frac{1-{{\left( \frac{1}{2} \right)}^{n+1}}}{1-\frac{1}{2}}.{{\log }_{a}}2017=\frac{{{2}^{n+1}}-1}{{{2}^{n}}}.{{\log }_{a}}2017\)

\(VP=2{{\log }_{a}}2017-\frac{1}{{{2}^{2018}}}.{{\log }_{a}}2017=\left( 2-\frac{1}{{{2}^{2018}}} \right).{{\log }_{a}}2017=\frac{{{2}^{2019}}-1}{{{2}^{2018}}}.{{\log }_{a}}2017\)

Do đó \(\frac{{{2}^{n+1}}-1}{{{2}^{n}}}=\frac{{{2}^{2019}}-1}{{{2}^{2018}}}\)\(\Leftrightarrow n=2018\).

Chọn B.