Tính:
LG a
\(A = {9^{2{{\log }_3}4 + 4{{\log }_{81}}2}}\)
Lời giải chi tiết:
\(\begin{array}{l}
A = {9^{2{{\log }_3}4 + 4{{\log }_{81}}2}} = {9^{{{\log }_3}{4^2} + 4{{\log }_{{3^4}}}2}}\\
= {9^{{{\log }_3}16 + \frac{4}{4}{{\log }_3}2}} = {9^{{{\log }_3}16 + {{\log }_3}2}}\\
= {9^{{{\log }_3}32}} = {\left( {{3^2}} \right)^{{{\log }_3}32}}\\
= {3^{2{{\log }_3}32}} = {\left( {{3^{{{\log }_3}32}}} \right)^2} = {32^2}\\
= 1024
\end{array}\)
LG b
\(B = {\log _a}\left( {{{{a^2}.\root 3 \of a .\root 5 \of {{a^4}} } \over {\root 4 \of a }}} \right)\)
Lời giải chi tiết:
Ta có:
\({{{a^2}.\root 3 \of a .\root 5 \of {{a^4}} } \over {\root 4 \of a }} = \frac{{{a^2}.{a^{\frac{1}{3}}}.{a^{\frac{4}{5}}}}}{{{a^{\frac{1}{4}}}}}\) \( = {a^{2 + {1 \over 3} + {4 \over 5} - {1 \over 4}}} = {a^{{{173} \over {60}}}}\)
Do đó: \(B = {\log _a}{a^{{{173} \over {60}}}} = {{173} \over {60}}\)
LG c
\(C = {\log _5}{\log _5}\root 5 \of {\root 5 \of {\root 5 \of {....\root 5 \of 5 } } } \)
Lời giải chi tiết:
\(\begin{array}{l}
\sqrt[5]{5} = {5^{\frac{1}{5}}}\\
\sqrt[5]{{\sqrt[5]{5}}} = {\left( {{5^{\frac{1}{5}}}} \right)^{\frac{1}{5}}} = {5^{\frac{1}{5}.\frac{1}{5}}} = {5^{{{\left( {\frac{1}{5}} \right)}^2}}}\\
...\\
\sqrt[5]{{\sqrt[5]{{\sqrt[5]{{...\sqrt[5]{5}}}}}}} = {5^{{{\left( {\frac{1}{5}} \right)}^n}}}\\
\Rightarrow {\log _5}{\log _5}\sqrt[5]{{\sqrt[5]{{\sqrt[5]{{...\sqrt[5]{5}}}}}}}\\
= {\log _5}{\log _5}\left[ {{5^{{{\left( {\frac{1}{5}} \right)}^n}}}} \right]\\
= {\log _5}{\left( {\frac{1}{5}} \right)^n} = {\log _5}{5^{ - n}} = - n\\
\Rightarrow C = - n
\end{array}\)
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