Với \(a\) là số thực dương tùy ý, \({\log _2}{a^2} + {\log _4}a\) bằng

Ta có \({4^x} + {4^{ - x}} = 14\)\( \Leftrightarrow {\left( {{2^x}} \right)^2} + {\left( {{2^{ - x}}} \right)^2} + 2 = 16\)\( \Leftrightarrow {\left( {{2^x} + {2^{ - x}}} \right)^2} = 16\)\( \Leftrightarrow \left[ \begin{array}{l}{2^x} + {2^{ - x}} = 4\\{2^x} + {2^{ - x}} =  - 4\end{array} \right.\)\( \Leftrightarrow {2^x} + {2^{ - x}} = 4\) (vì \({2^x} + {2^{ - x}} > 0,\forall x \in \mathbb{R}\)).

Vậy \(P = 4\).

Ta có: \(2 = {\log _{{a^3}}}\frac{{{a^5}}}{{\sqrt[4]{b}}} = \frac{1}{3}{\log _a}\frac{{{a^5}}}{{{b^{\frac{1}{4}}}}}\)\( = \frac{1}{3}\left( {{{\log }_a}{a^5} - {{\log }_a}{b^{\frac{1}{4}}}} \right)\)\( = \frac{1}{3}\left( {5 - \frac{1}{4}{{\log }_a}b} \right)\)

\( \Rightarrow 5 - \frac{1}{4}{\log _a}b = 6\)\( \Rightarrow {\log _a}b =  - 4\).