Ta có l o g a b bằng

Đặt \(t = {\rm{lo}}{{\rm{g}}_a}b > 0\). Suy ra \({\log _b}a = \frac{1}{{{{\log }_a}b}} = \frac{1}{t}\).

Ta có \(\frac{1}{{{\rm{lo}}{{\rm{g}}_{ab}}a}} + \frac{1}{{{\rm{lo}}{{\rm{g}}_{\sqrt[4]{{ab}}}}b}} = \frac{9}{4}\)\( \Leftrightarrow 1 + {\rm{lo}}{{\rm{g}}_a}b + \frac{1}{4}\left( {1 + {\rm{lo}}{{\rm{g}}_b}a} \right) = \frac{9}{4}\)\( \Leftrightarrow 1 + t + \frac{1}{4} + \frac{1}{{4t}} = \frac{9}{4}\)

\( \Leftrightarrow 4t + 4{t^2} + t + 1 = 9t \Leftrightarrow 4{t^2} - 4t + 1 = 0 \Leftrightarrow t = \frac{1}{2}\).

Vậy \({\rm{lo}}{{\rm{g}}_a}b = \frac{1}{2}\). Chọn B.