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Câu hỏi: Cho $a={{\log }_{2}}5,b={{\log }_{2}}9.$ Biểu diễn ${{\log }_{2}}\dfrac{40}{3}$ theo $a,b.$
A. $\dfrac{3a}{2b}.$
B. $3+a-2b.$
C. $3+a-\sqrt{b}.$
D. $3+a-\dfrac{b}{2}.$
A. $\dfrac{3a}{2b}.$
B. $3+a-2b.$
C. $3+a-\sqrt{b}.$
D. $3+a-\dfrac{b}{2}.$
Ta có ${{\log }_{2}}\dfrac{40}{3}={{\log }_{2}}40-{{\log }_{2}}3={{\log }_{2}}\left( {{5.2}^{3}} \right)-\dfrac{1}{2}{{\log }_{2}}9$
$={{\log }_{2}}5+{{\log }_{2}}{{2}^{3}}-\dfrac{1}{2}{{\log }_{2}}9=3+{{\log }_{2}}5-\dfrac{1}{2}{{\log }_{2}}9=3+a-\dfrac{1}{2}b.$
$={{\log }_{2}}5+{{\log }_{2}}{{2}^{3}}-\dfrac{1}{2}{{\log }_{2}}9=3+{{\log }_{2}}5-\dfrac{1}{2}{{\log }_{2}}9=3+a-\dfrac{1}{2}b.$
Đáp án D.