2019 AMC 12A 第 23 题

先试着解答 2019 AMC 12A 第 23 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2019 AMC 12A 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

23.

定义二元运算 ♢\diamondsuit 和 ♡\heartsuit 如下:
a♢b=alog⁡7(b) a \diamondsuit b = a^{\log_7(b)} 和 a♡b=a1log⁡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 aa 和 bb 成立。数列 (an)(a_n) 递归定义为 a3=3♡2a_3 = 3 \heartsuit 2,且 an=(n♡(n−1))♢an−1 a_n = (n \heartsuit (n - 1)) \diamondsuit a_{n-1} 对所有整数 n≥4n \ge 4 成立。最接近 log⁡7(a2019)\log_7(a_{2019}) 的整数是多少?
Define binary operations ♢\diamondsuit and ♡\heartsuit by
a♢b=alog⁡7(b) a \diamondsuit b = a^{\log_7(b)} and a♡b=a1log⁡7(b) a \heartsuit b = a^{\frac{1}{\log_7(b)}}
for all real numbers aa and bb for which these expressions are defined. The sequence (an)(a_n) is defined recursively by a3=3♡2a_3 = 3 \heartsuit 2 and an=(n♡(n−1))♢an−1 a_n = (n \heartsuit (n - 1)) \diamondsuit a_{n-1} for all integers n≥4.n \ge 4. To the nearest integer, what is log⁡7(a2019)?\log_7(a_{2019})?
88
99
1010
1111
1212
答案:D
知识点:自定义运算对数裂项相消
小提示:
施加 log⁡7\log_7:则 log⁡7(a♢b)=log⁡7a⋅log⁡7b\log_7(a \diamondsuit b) = \log_7 a \cdot \log_7 b
Apply log⁡7:\log_7: then log⁡7(a♢b)=log⁡7a⋅log⁡7b\log_7(a \diamondsuit b) = \log_7 a \cdot \log_7 b
大提示:
记 L(x)=log⁡7xL(x) = \log_7 x,递推变为 L(an)=L(an−1)⋅L(n)L(n−1)L(a_n) = L(a_{n-1}) \cdot \dfrac{L(n)}{L(n-1)}
Writing L(x)=log⁡7x,L(x) = \log_7 x, the recurrence becomes L(an)=L(an−1)⋅L(n)L(n−1)L(a_n) = L(a_{n-1}) \cdot \dfrac{L(n)}{L(n-1)}
解答:
令 L(x)=log⁡7xL(x) = \log_7 x。则 L(a♢b)=L(a)L(b)L(a \diamondsuit b) = L(a)L(b),且 L(a♡b)=L(a)L(b)L(a \heartsuit b) = \dfrac{L(a)}{L(b)}。
所以 L(a3)=L(3)L(2)L(a_3) = \dfrac{L(3)}{L(2)},且 L(an)=L(n)L(n−1)⋅L(an−1)L(a_n) = \dfrac{L(n)}{L(n-1)} \cdot L(a_{n-1})。该乘积会裂项相消:L(aN)=L(3)L(2)⋅L(N)L(3)=L(N)L(2)。 \begin{aligned} L(a_N) &= \dfrac{L(3)}{L(2)} \cdot \dfrac{L(N)}{L(3)} \\ &= \dfrac{L(N)}{L(2)} \end{aligned}\text{。}
因此 L(a2019)=log⁡72019log⁡72L(a_{2019}) = \dfrac{\log_7 2019}{\log_7 2} =log⁡22019≈10.98= \log_2 2019 \approx 10.98,四舍五入为 1111。
所以正确答案是 D。
Let L(x)=log⁡7x.L(x) = \log_7 x. Then L(a♢b)=L(a)L(b)L(a \diamondsuit b) = L(a)L(b) and L(a♡b)=L(a)L(b).L(a \heartsuit b) = \dfrac{L(a)}{L(b)}.
So L(a3)=L(3)L(2),L(a_3) = \dfrac{L(3)}{L(2)}, and L(an)=L(n)L(n−1)⋅L(an−1).L(a_n) = \dfrac{L(n)}{L(n-1)} \cdot L(a_{n-1}). The product telescopes: L(aN)=L(3)L(2)⋅L(N)L(3)=L(N)L(2). \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(a2019)=log⁡72019log⁡72L(a_{2019}) = \dfrac{\log_7 2019}{\log_7 2} =log⁡22019≈10.98,= \log_2 2019 \approx 10.98, which rounds to 11.11.
Thus, the correct answer is D.
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