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Câu hỏi: Cho ${{2}^{a}}=3,{{2}^{b}}=12$. Khi đó $a-b$ bằng
A. ${{\log }_{2}}36$.
B. $-2$.
C. $2$.
D. $-4$.
A. ${{\log }_{2}}36$.
B. $-2$.
C. $2$.
D. $-4$.
Ta có: ${{2}^{a}}=3\Leftrightarrow a={{\log }_{2}}3$, ${{2}^{b}}=12\Leftrightarrow b={{\log }_{2}}12$.
Khi đó $a-b={{\log }_{2}}3-{{\log }_{2}}12={{\log }_{2}}\dfrac{3}{12}={{\log }_{2}}\dfrac{1}{4}=-2$.
Khi đó $a-b={{\log }_{2}}3-{{\log }_{2}}12={{\log }_{2}}\dfrac{3}{12}={{\log }_{2}}\dfrac{1}{4}=-2$.
Đáp án B.