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Câu hỏi: [ Mức độ 2] Nếu ${{\log }_{7}}x=8{{\log }_{7}}a{{b}^{2}}-2{{\log }_{7}}{{a}^{3}}b \left( a,b>0 \right)$ thì $x$ bằng:
A. ${{a}^{8}}{{b}^{14}}$.
B. ${{a}^{6}}{{b}^{12}}$.
C. ${{a}^{6}}{{b}^{6}}$.
D. ${{\text{a}}^{2}}{{\text{b}}^{\text{14}}}$.
A. ${{a}^{8}}{{b}^{14}}$.
B. ${{a}^{6}}{{b}^{12}}$.
C. ${{a}^{6}}{{b}^{6}}$.
D. ${{\text{a}}^{2}}{{\text{b}}^{\text{14}}}$.
Ta có
$\begin{aligned}
& {{\log }_{7}}x=8{{\log }_{7}}a{{b}^{2}}-2{{\log }_{7}}{{a}^{3}}b \\
& \Leftrightarrow {{\log }_{7}}x={{\log }_{7}}{{a}^{8}}{{b}^{16}}-{{\log }_{7}}{{a}^{6}}{{b}^{2}} \\
& \Leftrightarrow {{\log }_{7}}x={{\log }_{7}}\dfrac{{{a}^{8}}{{b}^{16}}}{{{a}^{6}}{{b}^{2}}} \\
& \Leftrightarrow {{\log }_{7}}x={{\log }_{7}}{{a}^{2}}{{b}^{14}} \\
& \Leftrightarrow x={{a}^{2}}{{b}^{14}} \\
\end{aligned}$
$\begin{aligned}
& {{\log }_{7}}x=8{{\log }_{7}}a{{b}^{2}}-2{{\log }_{7}}{{a}^{3}}b \\
& \Leftrightarrow {{\log }_{7}}x={{\log }_{7}}{{a}^{8}}{{b}^{16}}-{{\log }_{7}}{{a}^{6}}{{b}^{2}} \\
& \Leftrightarrow {{\log }_{7}}x={{\log }_{7}}\dfrac{{{a}^{8}}{{b}^{16}}}{{{a}^{6}}{{b}^{2}}} \\
& \Leftrightarrow {{\log }_{7}}x={{\log }_{7}}{{a}^{2}}{{b}^{14}} \\
& \Leftrightarrow x={{a}^{2}}{{b}^{14}} \\
\end{aligned}$
Đáp án D.