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Câu hỏi: Cho ${{\log }_{6}}4=a$.Tính ${{\log }_{2}}3$ theo $a$.
A. $\dfrac{2-a}{a}$
B. $\dfrac{a+2}{a}$
C. $\dfrac{a-2}{a}$.
D. $\dfrac{a}{2-a}$.
A. $\dfrac{2-a}{a}$
B. $\dfrac{a+2}{a}$
C. $\dfrac{a-2}{a}$.
D. $\dfrac{a}{2-a}$.
Ta có: ${{\log }_{6}}4={{\log }_{2.3}}{{2}^{2}}=2.\dfrac{1}{{{\log }_{2}}\left( 2.3 \right)}=\dfrac{2}{1+{{\log }_{2}}3}=a$
$\Leftrightarrow {{\log }_{2}}3=\dfrac{2-a}{a}$.
$\Leftrightarrow {{\log }_{2}}3=\dfrac{2-a}{a}$.
Đáp án A.