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Câu hỏi: Cho hai số ${a, b}$ thỏa mãn ${\left( {{{\log }_2}a} \right).\left( {{{\log }_b}\sqrt 2 } \right) = 4}$. Tính ${\log _{ab}^{}a}$.
A. ${\dfrac{2}{3}}$.
B. ${\dfrac{8}{9}}$.
C. ${\dfrac{4}{3}}$.
D. ${\dfrac{{27}}{8}}$.
A. ${\dfrac{2}{3}}$.
B. ${\dfrac{8}{9}}$.
C. ${\dfrac{4}{3}}$.
D. ${\dfrac{{27}}{8}}$.
Điều kiện để các lôgarit có nghĩa: $\left\{ \begin{aligned}
& a>0 \\
& 0<b\ne 1 \\
& ab\ne 1 \\
\end{aligned} \right.$
Ta có: $\left( {{\log }_{2}}a \right).\left( {{\log }_{b}}\sqrt{2} \right)=4$
$\Leftrightarrow \left( {{\log }_{b}}{{2}^{\dfrac{1}{2}}} \right).\left( {{\log }_{2}}a \right)=4\Leftrightarrow \dfrac{1}{2}.\left( {{\log }_{b}}2 \right).\left( {{\log }_{2}}a \right)=4\Leftrightarrow \dfrac{1}{2}.{{\log }_{b}}a=4$ $\Leftrightarrow $ ${{\log }_{b}}a=8$
Khi đó ${{\log }_{ab}}a=\dfrac{{{\log }_{b}}a}{{{\log }_{b}}\left( ab \right)}=\dfrac{{{\log }_{b}}a}{{{\log }_{b}}a+{{\log }_{b}}b}=\dfrac{8}{8+1}=\dfrac{8}{9}$
& a>0 \\
& 0<b\ne 1 \\
& ab\ne 1 \\
\end{aligned} \right.$
Ta có: $\left( {{\log }_{2}}a \right).\left( {{\log }_{b}}\sqrt{2} \right)=4$
$\Leftrightarrow \left( {{\log }_{b}}{{2}^{\dfrac{1}{2}}} \right).\left( {{\log }_{2}}a \right)=4\Leftrightarrow \dfrac{1}{2}.\left( {{\log }_{b}}2 \right).\left( {{\log }_{2}}a \right)=4\Leftrightarrow \dfrac{1}{2}.{{\log }_{b}}a=4$ $\Leftrightarrow $ ${{\log }_{b}}a=8$
Khi đó ${{\log }_{ab}}a=\dfrac{{{\log }_{b}}a}{{{\log }_{b}}\left( ab \right)}=\dfrac{{{\log }_{b}}a}{{{\log }_{b}}a+{{\log }_{b}}b}=\dfrac{8}{8+1}=\dfrac{8}{9}$
Đáp án B.