Bài 29 trang 39 SBT Toán 11 - Cánh diều: Cho ${\log _a}b = 4.$ Tính: ${\log _a}\left( {{a^{\frac{1}{2}}}{b^5}} \right);$ ${\log _a}\left( {\frac{{a\sqrt b }}{{b\sqrt[3]{a}}}} \right);$ c) \({\log...

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) ${\log _{a\sqrt[3]{b}}}\left( {\sqrt[4]{{a\sqrt b }}} \right).$

Sử dụng các tính chất của logarit để tính giá trị biểu thức.

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: ${\log _a}b = 4 \Leftrightarrow b = {a^4}.$

$\Rightarrow {\log _{a\sqrt[3]{b}}}\left( {\sqrt[4]{{a\sqrt b }}} \right) = {\log _{a\sqrt[3]{{{a^4}}}}}{\left( {a\sqrt {{a^4}} } \right)^{\frac{1}{4}}} = {\log _{{a^{\frac{7}{3}}}}}{a^{\frac{3}{4}}} = \frac{3}{7}{\log _a}{a^{\frac{3}{4}}} = \frac{3}{7}.\frac{3}{4} = \frac{9}{{28}}.$