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12. (5 分) 设 $a=\log _{0.2} 0.3, b=\log _{2} 0.3$, 则 $(\quad)$

Gaokao · Math · previous-year question

  1. A.$a+b<a b<0$
  2. B.$a b<a+b<0$correct
  3. C.$a+b<0<a b$
  4. D.$a b<0<a+b$

Answer

B. $a b<a+b<0$

Explanation

解: $\because \mathrm{a}=\log _{0.2} 0.3=\frac{1 \mathrm{~g} 0.3}{-\lg 5}, \mathrm{~b}=\log _{2} 0.3=\frac{\lg 0.3}{1 \mathrm{~g} 2}$, $\therefore \mathrm{a}+\mathrm{b}=\frac{\lg 0.3}{\lg 2}-\frac{\lg 0.3}{\lg 5}=\frac{\lg 0.3(\lg 5-\lg 2)}{\lg 2 \lg 5}=\frac{\lg 0.3 \lg \frac{5}{2}}{\lg 2 \lg 5}$, $\mathrm{ab}=-\frac{\lg 0.3}{\lg 2} \cdot \frac{\lg 0.3}{\lg 5}=\frac{\lg 0.3 \cdot 1 \mathrm{~g} \frac{10}{3}}{\lg 2 \lg 5}$, $\because \lg \frac{10}{3}>\lg \frac{5}{2}, \frac{\lg 0.3}{\lg 2 \lg 5}<0$, $\therefore \mathrm{ab}<\mathrm{a}+\mathrm{b}<0$. 故选: B.

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