Google
Câu hỏi: Cho $a,b$ là các số dương. Tìm $x$ biết ${{\log }_{3}}x=4{{\log }_{3}}a+7{{\log }_{3}}b$.
A. $x={{a}^{\dfrac{1}{4}}}{{b}^{7}}.$
B. $x={{a}^{4}}{{b}^{\dfrac{1}{7}}}.$
C. $x={{a}^{4}}{{b}^{7}}$.
D. $x={{a}^{7}}{{b}^{4}}$.
A. $x={{a}^{\dfrac{1}{4}}}{{b}^{7}}.$
B. $x={{a}^{4}}{{b}^{\dfrac{1}{7}}}.$
C. $x={{a}^{4}}{{b}^{7}}$.
D. $x={{a}^{7}}{{b}^{4}}$.
Theo bài ra, ta có:
$\begin{aligned}
& {{\log }_{3}}x=4{{\log }_{3}}a+7{{\log }_{3}}b\Leftrightarrow {{\log }_{3}}x={{\log }_{3}}{{a}^{4}}+{{\log }_{3}}{{b}^{7}} \\
& \Leftrightarrow {{\log }_{3}}x={{\log }_{3}}{{a}^{4}}{{b}^{7}}\Leftrightarrow x={{a}^{4}}{{b}^{7}}. \\
\end{aligned}$
$\begin{aligned}
& {{\log }_{3}}x=4{{\log }_{3}}a+7{{\log }_{3}}b\Leftrightarrow {{\log }_{3}}x={{\log }_{3}}{{a}^{4}}+{{\log }_{3}}{{b}^{7}} \\
& \Leftrightarrow {{\log }_{3}}x={{\log }_{3}}{{a}^{4}}{{b}^{7}}\Leftrightarrow x={{a}^{4}}{{b}^{7}}. \\
\end{aligned}$
Đáp án C.