Question

定义二元运算 $\diamondsuit$ 和 $\heartsuit$ 如下：

和 $a \diamondsuit b = a^{\log_7(b)}$ $a \heartsuit b = a^{\frac{1}{\log_7(b)}}$

对所有使这些表达式有定义的实数 $a$ 和 $b$ 成立。数列 $(a_n)$ 递归定义为 $a_3 = 3 \heartsuit 2$ ，且对所有整数 $n \ge 4$ 成立。最接近 $\log_7(a_{2019})$ 的整数是多少？ $a_n = (n \heartsuit (n - 1)) \diamondsuit a_{n-1}$

Define binary operations $\diamondsuit$ and $\heartsuit$ by

$a \diamondsuit b = a^{\log_7(b)}$ and $a \heartsuit b = a^{\frac{1}{\log_7(b)}}$

for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3 \heartsuit 2$ and $a_n = (n \heartsuit (n - 1)) \diamondsuit a_{n-1}$ for all integers $n \ge 4.$ To the nearest integer, what is $\log_7(a_{2019})?$

$8$

$9$

$10$

$11$

$12$

答案：D

知识点：自定义运算对数裂项相消

难度评级：2240

解答：

令 $L(x) = \log_7 x$ 。则 $L(a \diamondsuit b) = L(a)L(b)$ ，且 $L(a \heartsuit b) = \dfrac{L(a)}{L(b)}$ 。

所以 $L(a_3) = \dfrac{L(3)}{L(2)}$ ，且 $L(a_n) = \dfrac{L(n)}{L(n-1)} \cdot L(a_{n-1})$ 。该乘积会望远镜相消： $\begin{aligned} L(a_N) &= \dfrac{L(3)}{L(2)} \cdot \dfrac{L(N)}{L(3)} \\ &= \dfrac{L(N)}{L(2)}. \end{aligned}$

因此 $L(a_{2019}) = \dfrac{\log_7 2019}{\log_7 2}$ $= \log_2 2019 \approx 10.98$ ，四舍五入为 $11$ 。

所以正确答案是 D。

Let $L(x) = \log_7 x.$ Then $L(a \diamondsuit b) = L(a)L(b)$ and $L(a \heartsuit b) = \dfrac{L(a)}{L(b)}.$

So $L(a_3) = \dfrac{L(3)}{L(2)},$ and $L(a_n) = \dfrac{L(n)}{L(n-1)} \cdot L(a_{n-1}).$ The product telescopes: $\begin{aligned} L(a_N) &= \dfrac{L(3)}{L(2)} \cdot \dfrac{L(N)}{L(3)} \\ &= \dfrac{L(N)}{L(2)}. \end{aligned}$

Hence $L(a_{2019}) = \dfrac{\log_7 2019}{\log_7 2}$ $= \log_2 2019 \approx 10.98,$ which rounds to $11.$

Thus, the correct answer is D.

2019 AMC 12A 第 23 题

答案：D

难度评级：2240

解答：

其他年份的第 23 题