Nội dung bài giảng
Bài 67. a) \({\log _2}x + {\log _4}x = {\log _{{1 \over 2}}}\sqrt 3 \);
b) \({\log _{\sqrt 3 }}x.{\log _3}x.{\log _9}x = 8\)
giải
a) Điều kiện: x > 0.
\({\log _2}x + {\log _4}x = {\log _{{1 \over 2}}}\sqrt 3 \Leftrightarrow {\log _2}x + {\log _{{2^2}}}x = {\log _{{2^{ - 1}}}}\sqrt 3 \)
\(\eqalign{
& \Leftrightarrow {\log _2}x + {1 \over 2}{\log _2}x = - {\log _2}\sqrt 3 \Leftrightarrow {3 \over 2}{\log _2}x = {\log _2}{1 \over {\sqrt 3 }} \cr
& \Leftrightarrow {\log _2}x = {\log _2}{\left( {{1 \over {\sqrt 3 }}} \right)^{{2 \over 3}}} \Leftrightarrow x = {1 \over {\root 3 \of 3 }} \cr} \)
Vậy \(S = \left\{ {1 \over{\root 3 \of 3 }} \right\}\)
b) Điều kiện: \(x > 0\).
\(\eqalign{
& {\log _{\sqrt 3 }}x.{\log _3}x.{\log _9}x = 8 \Leftrightarrow {\log _{{3^{{1 \over 2}}}}}x.{\log _3}x.{\log _{{3^2}}}x = 8 \cr
& \Leftrightarrow {1 \over {{1 \over 2}}}.{1 \over 2}.{\left( {{{\log }_3}x} \right)^3} = 8 \Leftrightarrow {\log _3}x = 2 \Leftrightarrow x = {3^2} = 9 \cr} \)
Vậy \(S = \left\{ 9 \right\}\)