Cho ${{\log }_{a}}x=2,{{\log }_{b}}x=3$ với $a,b$ là các số thực lớn hơn $1$. Tính $P={{\log }_{\dfrac{a}{{{b}^{2}}}}}x.$

Ta có $P={{\log }_{\dfrac{a}{{{b}^{2}}}}}x=\dfrac{1}{{{\log }_{x}}\dfrac{a}{{{b}^{2}}}}=\dfrac{1}{{{\log }_{x}}a-{{\log }_{x}}{{b}^{2}}}=\dfrac{1}{{{\log }_{x}}a-2{{\log }_{x}}b}$
Từ${{\log }_{a}}x=2,{{\log }_{b}}x=3\Rightarrow \left\{ \begin{aligned}
& {{\log }_{x}}a=\dfrac{1}{2} \\
& {{\log }_{x}}b=\dfrac{1}{3} \\
\end{aligned} \right.,$
Vậy $P={{\log }_{\dfrac{a}{{{b}^{2}}}}}x=\dfrac{1}{{{\log }_{x}}\dfrac{a}{{{b}^{2}}}}=\dfrac{1}{{{\log }_{x}}a-{{\log }_{x}}{{b}^{2}}}=\dfrac{1}{{{\log }_{x}}a-2{{\log }_{x}}b}=\dfrac{1}{\dfrac{1}{2}-2.\dfrac{1}{3}}=-6$.