Bài tập 29 trang 90 SGK Toán 12 NC

Áp dụng \({a^{lo{g_a}b}} = b(a > 0,a \ne 1)\)

\(\begin{array}{*{20}{l}}
{{3^{{{\log }_3}18}} = 18;{3^{5{{\log }_3}2}} = {3^{{{\log }_3}{2^5}}} = {2^5} = 32;}\\
\begin{array}{l}
{\left( {\frac{1}{8}} \right)^{{{\log }_2}5}} = {\left( {{2^{ - 3}}} \right)^{{{\log }_2}5}} = {2^{\left( { - 3} \right){{\log }_2}5}} = {2^{{{\log }_2}{5^{ - 3}}}}\\
 = {5^{ - 3}} = \frac{1}{{125}}
\end{array}\\
\begin{array}{l}
{\left( {\frac{1}{{32}}} \right)^{{{\log }_{0,5}}2}} = {\left( {{{\left( {\frac{1}{2}} \right)}^5}} \right)^{{{\log }_{\frac{1}{2}}}2}} = {\left( {\frac{1}{2}} \right)^{{{\log }_{\frac{1}{2}}}{2^5}}}\\
 = {2^5} = 32
\end{array}
\end{array}\)

-- Mod Toán 12 HỌC247