Giải bài 29 trang 90 SGK giải tích nâng cao 12

Hướng dẫn: ${{a}^{{{\log }_{a}}b}}={b}^{{{\log }_{a}}a}={{b}},\,\,\,\,\,\,\,\,\,\,\,\,\,{{a}^{n{{\log }_{a}}b}}={{b}^{n}}$

$\begin{align} & {{3}^{{{\log }_{3}}18}}={{18}^{{{\log }_{3}}3}}=18 \\ & {{3}^{5{{\log }_{3}}2}}={{3}^{{{\log }_{3}}{{2}^{5}}}}={{\left( {{2}^{5}} \right)}^{{{\log }_{3}}3}}=32 \\ & {{\left( \dfrac{1}{8} \right)}^{{{\log }_{2}}5}}={{\left( {{2}^{-3}} \right)}^{{{\log }_{2}}5}}={{\left( {{5}^{-3}} \right)}^{{{\log }_{2}}2}}=\dfrac{1}{125} \\ & {{\left( \dfrac{1}{32} \right)}^{{{\log }_{0,5}}2}}={{\left( {{2}^{-5}} \right)}^{{{\log }_{{{2}^{-1}}}}2}}={{\left( {{2}^{-5}} \right)}^{-1}}=32 \\ \end{align}$