Giải bài 86 trang 130 SGK giải tích nâng cao 12
Tính
a) \(A={{9}^{2{{\log }_{3}}4+4{{\log }_{81}}2}};\) b) \(B={{\log }_{a}}\left( \dfrac{{{a}^{2}}\sqrt[3]{a}\sqrt[5]{{{a}^{4}}}}{\sqrt[4]{a}} \right)\);
c) \(C={{\log }_{5}}{{\log }_{5}}\underbrace{\sqrt[5]{\sqrt[5]{\sqrt[5]{...\sqrt[5]{5}}}}}_{\text{n dấu căn}}\)
\(\begin{aligned} a)\, A&={{9}^{2{{\log }_{3}}4+4{{\log }_{81}}2}} \\ & ={{9}^{{{\log }_{3}}16+{{\log }_{3}}2}} \\ & ={{3}^{2{{\log }_{3}}32}}={{32}^{2{{\log }_{3}}3}} \\ & ={{32}^{2}}=1024 \\ b)\, B&={{\log }_{a}}\left( \dfrac{{{a}^{2}}\sqrt[3]{a}\sqrt[5]{{{a}^{4}}}}{\sqrt[4]{a}} \right) \\ & ={{\log }_{a}}{{a}^{2}}+{{\log }_{a}}\sqrt[3]{a}+{{\log }_{a}}\sqrt[5]{{{a}^{4}}}-{{\log }_{a}}\sqrt[4]{a} \\ & =2+\dfrac{1}{3}+\dfrac{4}{5}-\dfrac{1}{4} \\ & =\dfrac{173}{60} \\ c)\,& C={{\log }_{5}}{{\log }_{5}}\underbrace{\sqrt[5]{\sqrt[5]{\sqrt[5]{...\sqrt[5]{5}}}}}_{\text{n dấu căn}} \\ & \sqrt[5]{\sqrt[5]{\sqrt[5]{...\sqrt[5]{5}}}}={{5}^{{{\left( \frac{1}{5} \right)}^{n}}}} \\ & \Rightarrow {{\log }_{5}}{{\log }_{5}}\sqrt[5]{\sqrt[5]{\sqrt[5]{...\sqrt[5]{5}}}}={{\log }_{5}}{{\log }_{5}}{{5}^{{{\left( \frac{1}{5} \right)}^{n}}}} \\ & ={{\log }_{5}}{{\left( \dfrac{1}{5} \right)}^{n}}=-n \\ \end{aligned} \)