Bài 67 trang 124 SGK giải tích 12 nâng cao

LG a

\({\log _2}x + {\log _4}x = {\log _{{1 \over 2}}}\sqrt 3 \);

Lời giải chi tiết:

Điều kiện: x > 0.
\({\log _2}x + {\log _4}x = {\log _{{1 \over 2}}}\sqrt 3 \)

\(\Leftrightarrow {\log _2}x + {\log _{{2^2}}}x = {\log _{{2^{ - 1}}}}\sqrt 3 \)

\(\eqalign{
& \Leftrightarrow {\log _2}x + {1 \over 2}{\log _2}x = - {\log _2}\sqrt 3\cr& \Leftrightarrow {3 \over 2}{\log _2}x = {\log _2}{1 \over {\sqrt 3 }} \cr 
& \Leftrightarrow {\log _2}x = \frac{2}{3}{\log _2}\frac{1}{{\sqrt 3 }}\cr&\Leftrightarrow {\log _2}x = {\log _2}{\left( {{1 \over {\sqrt 3 }}} \right)^{{2 \over 3}}} \cr} \)

\(\begin{array}{l}
\Leftrightarrow x = {\left( {\frac{1}{{\sqrt 3 }}} \right)^{\frac{2}{3}}} = {\left( {{3^{ - \frac{1}{2}}}} \right)^{\frac{2}{3}}} = {3^{ - \frac{1}{3}}}\\
\Leftrightarrow x = \frac{1}{{\sqrt[3]{3}}}
\end{array}\)

Vậy \(S = \left\{ {1 \over{\root 3 \of 3 }} \right\}\)

LG b

\({\log _{\sqrt 3 }}x.{\log _3}x.{\log _9}x = 8\)

Lời giải chi tiết:

Điều kiện: \(x > 0\).

\(\eqalign{
& {\log _{\sqrt 3 }}x.{\log _3}x.{\log _9}x = 8 \cr&\Leftrightarrow {\log _{{3^{{1 \over 2}}}}}x.{\log _3}x.{\log _{{3^2}}}x = 8 \cr 
& \Leftrightarrow 2{\log _3}x.{\log _3}x.\frac{1}{2}{\log _3}x = 8\cr&\Leftrightarrow {\left( {{{\log }_3}x} \right)^3} = 8 \cr&\Leftrightarrow {\log _3}x = 2 \cr&\Leftrightarrow x = {3^2} = 9 \cr} \)

Vậy \(S = \left\{ 9 \right\}\)

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