Cho ${{\log }_{ab}}b=3$ (với $a>0,b>0,ab\ne 1$ ). Tính ${{\log...

Ta có ${{\log }_{ab}}b=3\Leftrightarrow {{\log }_{b}}ab=\dfrac{1}{3}\Leftrightarrow {{\log }_{b}}a+1=\dfrac{1}{3}\Leftrightarrow {{\log }_{b}}a=-\dfrac{2}{3}$
Ta có ${{\log }_{\sqrt{ab}}}\left( \dfrac{a}{{{b}^{2}}} \right)=2{{\log }_{ab}}\left( \dfrac{a}{{{b}^{2}}} \right)=2{{\log }_{ab}}a-4{{\log }_{ab}}b=\dfrac{2}{{{\log }_{a}}ab}-4.3=\dfrac{2}{1+{{\log }_{a}}b}-12=\dfrac{2}{1-\dfrac{3}{2}}-12=-16$