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Câu hỏi: Đơn giản các biểu thức:
Lời giải chi tiết:
\(\log {1 \over 8} + {1 \over 2}\log 4 + 4\log \sqrt 2 \)
\(\begin{array}{l}
= \log \frac{1}{8} + \log {4^{\frac{1}{2}}} + \log {\left({\sqrt 2 } \right)^4}\\
= \log \frac{1}{8} + \log \sqrt 4 + \log {\left[ {{{\left({\sqrt 2 } \right)}^2}} \right]^2}\\= \log \frac{1}{8} + \log 2 + \log 4\\
= \log \left({\frac{1}{8}. 2.4} \right)\\
= \log 1\\
= 0
\end{array}\)
Lời giải chi tiết:
\(\log {4 \over 9} + {1 \over 2}\log 36 + {3 \over 2}\log {9 \over 2}\)
\(\begin{array}{l}
= \log \frac{4}{9} + \log {36^{\frac{1}{2}}} + \log {\left({\frac{9}{2}} \right)^{\frac{3}{2}}}\\
= \log \frac{4}{9} + \log 6 + \log \left({\sqrt {{{\left( {\frac{9}{2}} \right)}^3}} } \right)\\
= \log \left({\frac{4}{9}. 6.\sqrt {{{\left( {\frac{9}{2}} \right)}^3}} } \right)\\
= \log \left({\frac{8}{3}.\frac{9}{2}\sqrt {\frac{9}{2}} } \right)\\
= \log \left({12.\frac{3}{{\sqrt 2 }}} \right)\\
= \log \left({18\sqrt 2 } \right)
\end{array}\)
Lời giải chi tiết:
$\begin{array}{l}
\log 72 - 2\left( {\log 27 - \log 256} \right) + \log {108^{\frac{1}{2}}}\\
= \log 72 - 2\left( {\log 27 - \log 256} \right) + \frac{1}{2}\log 108\\
= \log \left( {{2^3}{{.3}^2}} \right) - 2\left( {\log {3^2} - \log {2^8}} \right) + \frac{1}{2}\log \left( {{2^2}{{.3}^3}} \right)\\
= \log {2^3} + \log {3^2} - 2\log {3^3} + 2\log {2^8} + \frac{1}{2}\left( {\log {2^2} + \log {3^3}} \right)\\
= 3\log 2 + 2\log 3 - 6\log 3 + 16\log 2 + \frac{1}{2}\left( {2\log 2 + 3\log 3} \right)\\
= 19\log 2 - 4\log 3 + \log 2 + \frac{3}{2}\log 3\\
= 20\log 2 - \frac{5}{2}\log 3
\end{array}$
Lời giải chi tiết:
\(\begin{array}{l}
= \log \frac{1}{8} - \log \frac{3}{8} + \log {\left({\sqrt {0,5625} } \right)^2}\\
= \log \frac{1}{8} - \log \frac{3}{8} + \log 0,5625\\
= \log \frac{1}{8} - \log \frac{3}{8} + \log \frac{9}{{16}}\\
= \left({\log \frac{1}{8} + \log \frac{9}{{16}}} \right) - \log \frac{3}{8}\\
= \log \left({\frac{1}{8}.\frac{9}{{16}}} \right) - \log \frac{3}{8}\\
= \log \frac{9}{{128}} - \log \frac{3}{8}\\
= \log \left({\frac{9}{{128}}:\frac{3}{8}} \right)\\
= \log \frac{3}{{16}}
\end{array}\)
Câu a
\(\log {1 \over 8} + {1 \over 2}\log 4 + 4\log \sqrt 2 \)Lời giải chi tiết:
\(\log {1 \over 8} + {1 \over 2}\log 4 + 4\log \sqrt 2 \)
\(\begin{array}{l}
= \log \frac{1}{8} + \log {4^{\frac{1}{2}}} + \log {\left({\sqrt 2 } \right)^4}\\
= \log \frac{1}{8} + \log \sqrt 4 + \log {\left[ {{{\left({\sqrt 2 } \right)}^2}} \right]^2}\\= \log \frac{1}{8} + \log 2 + \log 4\\
= \log \left({\frac{1}{8}. 2.4} \right)\\
= \log 1\\
= 0
\end{array}\)
Câu b
\(\log {4 \over 9} + {1 \over 2}\log 36 + {3 \over 2}\log {9 \over 2}\)Lời giải chi tiết:
\(\log {4 \over 9} + {1 \over 2}\log 36 + {3 \over 2}\log {9 \over 2}\)
\(\begin{array}{l}
= \log \frac{4}{9} + \log {36^{\frac{1}{2}}} + \log {\left({\frac{9}{2}} \right)^{\frac{3}{2}}}\\
= \log \frac{4}{9} + \log 6 + \log \left({\sqrt {{{\left( {\frac{9}{2}} \right)}^3}} } \right)\\
= \log \left({\frac{4}{9}. 6.\sqrt {{{\left( {\frac{9}{2}} \right)}^3}} } \right)\\
= \log \left({\frac{8}{3}.\frac{9}{2}\sqrt {\frac{9}{2}} } \right)\\
= \log \left({12.\frac{3}{{\sqrt 2 }}} \right)\\
= \log \left({18\sqrt 2 } \right)
\end{array}\)
Câu c
\(\log 72 - 2\log {{27} \over {256}} + \log \sqrt {108} \)Lời giải chi tiết:
$\begin{array}{l}
\log 72 - 2\left( {\log 27 - \log 256} \right) + \log {108^{\frac{1}{2}}}\\
= \log 72 - 2\left( {\log 27 - \log 256} \right) + \frac{1}{2}\log 108\\
= \log \left( {{2^3}{{.3}^2}} \right) - 2\left( {\log {3^2} - \log {2^8}} \right) + \frac{1}{2}\log \left( {{2^2}{{.3}^3}} \right)\\
= \log {2^3} + \log {3^2} - 2\log {3^3} + 2\log {2^8} + \frac{1}{2}\left( {\log {2^2} + \log {3^3}} \right)\\
= 3\log 2 + 2\log 3 - 6\log 3 + 16\log 2 + \frac{1}{2}\left( {2\log 2 + 3\log 3} \right)\\
= 19\log 2 - 4\log 3 + \log 2 + \frac{3}{2}\log 3\\
= 20\log 2 - \frac{5}{2}\log 3
\end{array}$
Câu d
\(\log {1 \over 8} - \log 0,375 + 2\log \sqrt {0,5625} \)Lời giải chi tiết:
\(\begin{array}{l}
= \log \frac{1}{8} - \log \frac{3}{8} + \log {\left({\sqrt {0,5625} } \right)^2}\\
= \log \frac{1}{8} - \log \frac{3}{8} + \log 0,5625\\
= \log \frac{1}{8} - \log \frac{3}{8} + \log \frac{9}{{16}}\\
= \left({\log \frac{1}{8} + \log \frac{9}{{16}}} \right) - \log \frac{3}{8}\\
= \log \left({\frac{1}{8}.\frac{9}{{16}}} \right) - \log \frac{3}{8}\\
= \log \frac{9}{{128}} - \log \frac{3}{8}\\
= \log \left({\frac{9}{{128}}:\frac{3}{8}} \right)\\
= \log \frac{3}{{16}}
\end{array}\)
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