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Câu hỏi: Biết rằng $a={{\log }_{2}}3,b={{\log }_{2}}5$. Hãy biểu diễn ${{\log }_{45}}4$ theo $a$ và $b$
A. $\dfrac{a+2b}{2}$.
B. $2ab$.
C. $\dfrac{2}{2a+b}$.
D. $\dfrac{2a+b}{2}$.
A. $\dfrac{a+2b}{2}$.
B. $2ab$.
C. $\dfrac{2}{2a+b}$.
D. $\dfrac{2a+b}{2}$.
Ta có:
${{\log }_{45}}4=\dfrac{{{\log }_{2}}4}{{{\log }_{2}}45}=\dfrac{2{{\log }_{2}}2}{{{\log }_{2}}\left( {{5.3}^{2}} \right)}=\dfrac{2}{{{\log }_{2}}{{3}^{2}}+{{\log }_{2}}5}=\dfrac{2}{2{{\log }_{2}}3+{{\log }_{2}}5}=\dfrac{2}{2a+b}$
${{\log }_{45}}4=\dfrac{{{\log }_{2}}4}{{{\log }_{2}}45}=\dfrac{2{{\log }_{2}}2}{{{\log }_{2}}\left( {{5.3}^{2}} \right)}=\dfrac{2}{{{\log }_{2}}{{3}^{2}}+{{\log }_{2}}5}=\dfrac{2}{2{{\log }_{2}}3+{{\log }_{2}}5}=\dfrac{2}{2a+b}$
Đáp án C.