Bài tập 32 trang 92 SGK Toán 12 NC

Bài tập 32 trang 92 SGK Toán 12 NC

a)

\(\begin{array}{l}
{\log _8}12 - {\log _8}15 + {\log _8}20\\
 = {\log _8}\frac{{12.20}}{{15}} = {\log _8}16\\
 = {\log _{{2^3}}}{2^4} = \frac{4}{3}
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\frac{1}{2}{\log _7}36 - {\log _7}14 - 3{\log _7}\sqrt[3]{3}\\
 = {\log _7}6 - {\log _7}14 - {\log _7}21
\end{array}\\
\begin{array}{l}
 = {\log _7}\frac{6}{{14.21}} = {\log _7}\frac{1}{{49}}\\
 = {\log _7}{7^{ - 2}} =  - 2
\end{array}
\end{array}\)

c)

\(\begin{array}{l}
\frac{{{{\log }_5}36 - {{\log }_5}12}}{{{{\log }_5}9}}\\
 = \frac{{{{\log }_5}\frac{{36}}{{12}}}}{{{{\log }_5}{3^2}}} = \frac{{{{\log }_5}3}}{{2{{\log }_5}3}} = \frac{1}{2}
\end{array}\)

d)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{36^{{{\log }_6}5}} + {10^{1 - \log 2}} - {8^{{{\log }_2}3}}\\
 = {6^{2{{\log }_6}5}} + {10^{{{\log }_{10}}\frac{{10}}{2}}} - {2^{{{\log }_2}27}}
\end{array}\\
\begin{array}{l}
 = {6^{\log {{65}^2}}} + {10^{{{\log }_{10}}5}} - {2^{{{\log }_2}27}}\\
 = 25 + 5 - 27 = 3
\end{array}
\end{array}\)

-- Mod Toán 12