Với a, b > 0, cho ({log _{a{b^{ - 3}}}}a = frac{1}{4})

Ta có: \({\log _{a{b^{ - 3}}}}a = \frac{1}{{1 - 3{{\log }_a}b}} = \frac{1}{4} \Rightarrow {\log _a}b =  - 1.\)

Suy ra:

\({\log _{{a^3}b}}\sqrt {\frac{{{a^5}}}{b}}  = \frac{5}{2}{\log _{{a^3}b}}a - \frac{1}{2}{\log _{{a^3}b}}b = \frac{5}{2}.\frac{1}{{3 + {{\log }_a}b}} - \frac{1}{2}.\frac{1}{{1 + 3{{\log }_b}a}} = \frac{5}{2}.\frac{1}{{3 - 1}} - \frac{1}{2}.\frac{1}{{1 - 3}} = \frac{3}{2}.\)