AMC12 2020 B

AMC12 2020 B · Q13

AMC12 2020 B · Q13. It mainly tests Logarithms (rare), Manipulating equations.

Which of the following is equal to $\sqrt{\log_2 6 + \log_3 6}$?

下面哪一项等于$\sqrt{\log_2 6 + \log_3 6}$？

(A) 1 1

(B) $\sqrt{\log_5 6}$ $\sqrt{\log_5 6}$

(D) $\sqrt{\log_2 3 + \sqrt{\log_3 2}}$ $\sqrt{\log_2 3 + \sqrt{\log_3 2}}$

(E) $\sqrt{\log_2 6 + \sqrt{\log_3 6}}$ $\sqrt{\log_2 6 + \sqrt{\log_3 6}}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $a=\sqrt{\log_2 3}$. Note that $\frac{1}{a}=\sqrt{\log_3 2}$. Then \[ \sqrt{\log_2 6+\log_3 6} =\sqrt{\log_2 3+\log_3 2+2} =\sqrt{a^2+2+\frac{1}{a^2}} =a+\frac{1}{a} =\sqrt{\log_2 3}+\sqrt{\log_3 2}. \] It remains to show that the other choices given are not equal to $\sqrt{\log_2 6+\log_3 6}$, and to do so it is enough to show that the choices are in strictly increasing order. Because $6>5$ it follows that $1<\sqrt{\log_5 6}$. Because $6<25$ it follows that $\sqrt{\log_5 6}<2$. The Arithmetic Mean-Geometric Mean Inequality and the fact that $\log_2 3\ne \log_3 2$ imply that \[ \frac{\sqrt{\log_2 3}+\sqrt{\log_3 2}}{2}>\sqrt{\sqrt{\log_2 3}\cdot \sqrt{\log_3 2}}=1, \] from which it follows that $2<\sqrt{\log_2 3}+\sqrt{\log_3 2}$. Finally each term of choice (D) is less than the corresponding term in choice (E), so the sum is less as well.

答案（D）：令 $a=\sqrt{\log_2 3}$。注意到 $\frac{1}{a}=\sqrt{\log_3 2}$。则 \[ \sqrt{\log_2 6+\log_3 6} =\sqrt{\log_2 3+\log_3 2+2} =\sqrt{a^2+2+\frac{1}{a^2}} =a+\frac{1}{a} =\sqrt{\log_2 3}+\sqrt{\log_3 2}. \] 还需说明其余选项不等于 $\sqrt{\log_2 6+\log_3 6}$；为此只需证明这些选项严格递增。因为 $6>5$，所以 $1<\sqrt{\log_5 6}$。因为 $6<25$，所以 $\sqrt{\log_5 6}<2$。由算术平均-几何平均不等式以及 $\log_2 3\ne \log_3 2$ 可得 \[ \frac{\sqrt{\log_2 3}+\sqrt{\log_3 2}}{2}>\sqrt{\sqrt{\log_2 3}\cdot \sqrt{\log_3 2}}=1, \] 从而得到 $2<\sqrt{\log_2 3}+\sqrt{\log_3 2}$。最后，选项（D）的每一项都小于选项（E）中对应的项，因此它们的和也更小。

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