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Câu hỏi: Cho $a>0,b>0,$ nếu viết ${{\log }_{3}}{{\left( \sqrt[5]{{{a}^{3}}b} \right)}^{\dfrac{2}{3}}}=\dfrac{x}{5}{{\log }_{3}}a+\dfrac{y}{15}{{\log }_{3}}b$ thì $x+y$ bằng bao nhiêu?
A. 5.
B. 2.
C. 4.
D. 3.
A. 5.
B. 2.
C. 4.
D. 3.
Ta có ${{\log }_{3}}{{\left( \sqrt[5]{{{a}^{3}}b} \right)}^{\dfrac{2}{3}}}=\dfrac{2}{3}.\dfrac{1}{5}{{\log }_{3}}\left( {{a}^{3}}b \right)$
$=\dfrac{2}{15}\left( {{\log }_{3}}{{a}^{3}}+{{\log }_{3}}b \right)$
$=\dfrac{2}{15}.3.{{\log }_{3}}a+\dfrac{2}{15}{{\log }_{3}}b$
$=\dfrac{2}{5}{{\log }_{3}}a+\dfrac{2}{15}{{\log }_{3}}b.$
Vậy $x=2,y=2\Rightarrow x+y=4.$
$=\dfrac{2}{15}\left( {{\log }_{3}}{{a}^{3}}+{{\log }_{3}}b \right)$
$=\dfrac{2}{15}.3.{{\log }_{3}}a+\dfrac{2}{15}{{\log }_{3}}b$
$=\dfrac{2}{5}{{\log }_{3}}a+\dfrac{2}{15}{{\log }_{3}}b.$
Vậy $x=2,y=2\Rightarrow x+y=4.$
Đáp án C.