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Câu hỏi: Cho $a>0$ và $a\ne 1$, khi đó ${{\log }_{{{a}^{3}}}}\left( 3a \right)$ bằng
A. $1$.
B. $\dfrac{1}{3}\left( 1+{{\log }_{a}}3 \right)$.
C. $3\left( 1+{{\log }_{a}}3 \right)$.
D. $-1+{{\log }_{a}}3$.
A. $1$.
B. $\dfrac{1}{3}\left( 1+{{\log }_{a}}3 \right)$.
C. $3\left( 1+{{\log }_{a}}3 \right)$.
D. $-1+{{\log }_{a}}3$.
${{\log }_{{{a}^{3}}}}\left( 3a \right)=\dfrac{1}{3}\left( {{\log }_{a}}3+{{\log }_{a}}a \right)=\dfrac{1}{3}\left( {{\log }_{a}}3+1 \right)$.
Đáp án B.