Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
23,\\
lo{g_a}b = \sqrt 5  \Rightarrow {\log _b}a = \dfrac{1}{{\sqrt 5 }}\\
P = {\log _{\dfrac{a}{{\sqrt b }}}}\sqrt {ab}  = {\log _{\dfrac{a}{{\sqrt b }}}}{\left( {ab} \right)^{\dfrac{1}{2}}} = \dfrac{1}{2}{\log _{\dfrac{a}{{\sqrt b }}}}ab\\
 = \dfrac{1}{2}\left( {{{\log }_{\dfrac{a}{{\sqrt b }}}}a + {{\log }_{\dfrac{a}{{\sqrt b }}}}b} \right) = \dfrac{1}{2}.\left( {\dfrac{1}{{{{\log }_a}\dfrac{a}{{\sqrt b }}}} + \dfrac{1}{{{{\log }_b}\dfrac{a}{{\sqrt b }}}}} \right)\\
 = \dfrac{1}{2}.\left( {\dfrac{1}{{{{\log }_a}a - {{\log }_a}\sqrt b }} + \dfrac{1}{{{{\log }_b}a - {{\log }_b}\sqrt b }}} \right)\\
 = \dfrac{1}{2}.\left( {\dfrac{1}{{1 - \dfrac{1}{2}{{\log }_a}b}} + \dfrac{1}{{{{\log }_b}a - \dfrac{1}{2}{{\log }_b}b}}} \right)\\
 = \dfrac{1}{2}.\left( {\dfrac{1}{{1 - \dfrac{1}{2}.\sqrt 5 }} + \dfrac{1}{{\dfrac{1}{{\sqrt 5 }} - \dfrac{1}{2}.1}}} \right)\\
 = \dfrac{1}{2}.\left( {\dfrac{1}{{\dfrac{{2 - \sqrt 5 }}{2}}} + \dfrac{1}{{\dfrac{{2 - \sqrt 5 }}{{2\sqrt 5 }}}}} \right)\\
 = \dfrac{1}{2}.\left( {\dfrac{2}{{2 - \sqrt 5 }} + \dfrac{{2\sqrt 5 }}{{2 - \sqrt 5 }}} \right) = \dfrac{{1 + \sqrt 5 }}{{2 - \sqrt 5 }}\\
 = \dfrac{{\left( {1 + \sqrt 5 } \right)\left( {2 + \sqrt 5 } \right)}}{{\left( {2 - \sqrt 5 } \right)\left( {2 + \sqrt 5 } \right)}} = \dfrac{{7 + 3\sqrt 5 }}{{ - 1}} =  - 7 - 3\sqrt 5 \\
24,\,\,\,\,\,\,lo{g_a}b = \sqrt 2  \Rightarrow {\log _b}a = \dfrac{1}{{\sqrt 2 }}\\
P = {\log _{{a^2}b}}\dfrac{{{b^2}}}{{\sqrt a }} = {\log _{{a^2}b}}{b^2} - {\log _{{a^2}b}}\sqrt a \\
 = 2{\log _{{a^2}b}}b - \dfrac{1}{2}{\log _{{a^2}b}}a = \dfrac{2}{{{{\log }_b}\left( {{a^2}b} \right)}} - \dfrac{1}{{2{{\log }_a}\left( {{a^2}b} \right)}}\\
 = \dfrac{2}{{{{\log }_b}{a^2} + {{\log }_b}b}} - \dfrac{1}{{2.\left( {{{\log }_a}{a^2} + {{\log }_a}b} \right)}}\\
 = \dfrac{2}{{2{{\log }_b}a + 1}} - \dfrac{1}{{2.\left( {2{{\log }_a}a + {{\log }_a}b} \right)}}\\
 = \dfrac{2}{{2.\dfrac{1}{{\sqrt 2 }} + 1}} - \dfrac{1}{{2.\left( {2.1 + \sqrt 2 } \right)}} = \dfrac{2}{{\sqrt 2  + 1}} - \dfrac{1}{{2.\left( {2 + \sqrt 2 } \right)}}\\
 = \dfrac{2}{{\sqrt 2  + 1}} - \dfrac{1}{{2\sqrt 2 \left( {\sqrt 2  + 1} \right)}} = \dfrac{{2.2\sqrt 2  - 1}}{{2\sqrt 2 \left( {\sqrt 2  + 1} \right)}}\\
 = \dfrac{{4\sqrt 2  - 1}}{{2\sqrt 2 \left( {\sqrt 2  + 1} \right)}} = \dfrac{{4\sqrt 2  - 1}}{{4 + 2\sqrt 2 }}
\end{array}\)