Đề bài

Cho \({\log _a}b = 4.\) Tính: a) \({\log _a}\left( {{a^{\frac{1}{2}}}{b^5}} \right);\)                                      b) \({\log _a}\left( {\frac{{a\sqrt b }}{{b\sqrt[3]{a}}}} \right);\) c) \({\log _{{a^3}{b^2}}}\left( {{a^2}{b^3}} \right);\)                                     d) \({\log _{a\sqrt[3]{b}}}\left( {\sqrt[4]{{a\sqrt b }}} \right).\)

Lời giải chi tiết

a) \({\log _a}\left( {{a^{\frac{1}{2}}}{b^5}} \right) = {\log _a}{a^{\frac{1}{2}}} + {\log _a}{b^5} = \frac{1}{2} + 5{\log _a}b = \frac{1}{2} + 5.4 = \frac{{41}}{2}.\)    b) \({\log _a}\left( {\frac{{a\sqrt b }}{{b\sqrt[3]{a}}}} \right) = {\log _a}\left( {\frac{{a{b^{\frac{1}{2}}}}}{{{a^{\frac{1}{3}}}b}}} \right) = {\log _a}\left( {{a^{\frac{2}{3}}}{b^{ - \frac{1}{2}}}} \right) = {\log _a}{a^{\frac{2}{3}}} + {\log _a}{b^{ - \frac{1}{2}}}\) \( = \frac{2}{3} - \frac{1}{2}{\log _a}b = \frac{2}{3} - \frac{1}{2}.4 =  - \frac{4}{3}.\)c) Ta có: \({\log _a}b = 4 \Leftrightarrow b = {a^4}.\)\( \Rightarrow {\log _{{a^3}{b^2}}}\left( {{a^2}{b^3}} \right) = {\log _{{a^3}.{a^8}}}\left( {{a^2}.{a^{12}}} \right) = {\log _{{a^{11}}}}{a^{14}} = \frac{1}{{11}}{\log _a}{a^{14}} = \frac{{14}}{{11}}.\)             d) Ta có: \({\log _a}b = 4 \Leftrightarrow b = {a^4}.\)\( \Rightarrow {\log _{a\sqrt[3]{b}}}\left( {\sqrt[4]{{a\sqrt b }}} \right) = {\log _{a\sqrt[3]{{{a^4}}}}}{\left( {a\sqrt {{a^4}} } \right)^{\frac{1}{4}}} = {\log _{{a^{\frac{7}{3}}}}}{a^{\frac{3}{4}}} = \frac{3}{7}{\log _a}{a^{\frac{3}{4}}} = \frac{3}{7}.\frac{3}{4} = \frac{9}{{28}}.\)