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Câu hỏi: Với mọi $a,b$ là các số thực dương, khác $1$ thỏa mãn ${{\log }_{{{a}^{2}}}}\left( {{a}^{16}}{{b}^{2}} \right)={{\log }_{\sqrt{a}}}\left( \dfrac{b}{\sqrt{a}} \right)$. Mệnh đề nào sau đây đúng?
A. ${{a}^{5}}=b.$
B. ${{a}^{2}}=b.$
C. ${{a}^{9}}=b.$
D. $a=b.$
Ta có: ${{\log }_{{{a}^{2}}}}\left( {{a}^{16}}{{b}^{2}} \right)={{\log }_{\sqrt{a}}}\left( \dfrac{b}{\sqrt{a}} \right)$ $\Leftrightarrow \dfrac{1}{2}{{\log }_{a}}\left( {{a}^{16}}{{b}^{2}} \right)=2{{\log }_{a}}\left( \dfrac{b}{\sqrt{a}} \right)$
$\Leftrightarrow {{\left( {{a}^{16}}{{b}^{2}} \right)}^{\dfrac{1}{2}}}={{\left( \dfrac{b}{\sqrt{a}} \right)}^{2}}$ $\Leftrightarrow {{a}^{8}}b=\dfrac{{{b}^{2}}}{a}\Leftrightarrow {{a}^{9}}=b$
A. ${{a}^{5}}=b.$
B. ${{a}^{2}}=b.$
C. ${{a}^{9}}=b.$
D. $a=b.$
Ta có: ${{\log }_{{{a}^{2}}}}\left( {{a}^{16}}{{b}^{2}} \right)={{\log }_{\sqrt{a}}}\left( \dfrac{b}{\sqrt{a}} \right)$ $\Leftrightarrow \dfrac{1}{2}{{\log }_{a}}\left( {{a}^{16}}{{b}^{2}} \right)=2{{\log }_{a}}\left( \dfrac{b}{\sqrt{a}} \right)$
$\Leftrightarrow {{\left( {{a}^{16}}{{b}^{2}} \right)}^{\dfrac{1}{2}}}={{\left( \dfrac{b}{\sqrt{a}} \right)}^{2}}$ $\Leftrightarrow {{a}^{8}}b=\dfrac{{{b}^{2}}}{a}\Leftrightarrow {{a}^{9}}=b$
Đáp án C.