- LG a
- LG b
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 \[S = \left\{ 9 \right\}\]