AMC12 2019 A

AMC12 2019 A · Q23

AMC12 2019 A · Q23. It mainly tests Logarithms (rare).

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

定义二元运算 $\diamond$ 和 $\diamond$ 如下： $$ a \diamond b = a^{\log_7(b)} $$ 和 $$ a \diamond b = a^{\frac{1}{\log_7(b)}} $$ 对所有这些表达式有定义的实数 $a$ 和 $b$。序列 $(a_n)$ 由 $a_3 = 3 \diamond 2$ 和 $$ a_n = (n \diamond (n-1)) \diamond a_{n-1} $$ 对所有整数 $n \ge 4$ 递归定义。四舍五入到最近整数，$\log_7(a_{2019})$ 是多少？

(A) 8 8

(B) 9 9

(D) 11 11

(E) 12 12

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Note that $a \diamond b = a^{\log_7(b)} = 7^{\log_7(a)\log_7(b)}.$ It follows that $\diamond$ operates with closure on the interval $(1,\infty)$ and is commutative and associative. Furthermore, the identity element for $\diamond$ is $7$, and the inverse of $a$ is $7^{\frac{1}{\log_7(a)}}$ for all $a>1$; denote the inverse of $a$ under $\diamond$ by $a^{-1}$. Note further that $a \heartsuit b = a^{\frac{1}{\log_7(b)}} = 7^{\log_7(a)\log_7\!\left(7^{\frac{1}{\log_7(b)}}\right)} = a \diamond b^{-1}.$ $s=\dfrac{80}{\sqrt{13}}-\dfrac{34}{\sqrt{3}}.$ Therefore the requested answer is $80+13+34+3=130$. Thus $\begin{aligned} \log_7(a_{2019}) &= \log_7((2019 \heartsuit 2018)\diamond\cdots\diamond(4\heartsuit 3)\diamond(3\heartsuit 2))\\ &= \log_7((2019\diamond 2018^{-1})\diamond\cdots\diamond(4\diamond 3^{-1})\diamond(3\diamond 2^{-1}))\\ &= \log_7(2019\diamond(2018^{-1}\diamond 2018)\diamond\cdots\diamond(3^{-1}\diamond 3)\diamond 2^{-1})\\ &= \log_7(2019\diamond 7\diamond\cdots\diamond 7\diamond 2^{-1})\\ &= \log_7(2019\diamond 2^{-1})\\ &= \log_7(2019^{\frac{1}{\log_7(2)}})\\ &= \dfrac{\log_7(2019)}{\log_7(2)}\\ &= \log_2(2019). \end{aligned}$ Because $2^{11}=2048$, the closest integer to $\log_2(2019)$ is $11$. (More precisely, $\log_2(2019)<\log_2(2048)=11$, and $2^{10.5}=2^{10}\cdot\sqrt{2}<1200\cdot 1.5=1800,$ so $\log_2(2019)>\log_2(1800)>10.5.)

答案（D）：注意 $a \diamond b = a^{\log_7(b)} = 7^{\log_7(a)\log_7(b)}.$ 由此可知，$\diamond$ 在区间 $(1,\infty)$ 上封闭，且满足交换律与结合律。此外，$\diamond$ 的单位元是 $7$，并且对所有 $a>1$，$a$ 的逆元为 $7^{\frac{1}{\log_7(a)}}$；用 $a^{-1}$ 表示 $a$ 在 $\diamond$ 下的逆元。还要注意 $a \heartsuit b = a^{\frac{1}{\log_7(b)}} = 7^{\log_7(a)\log_7\!\left(7^{\frac{1}{\log_7(b)}}\right)} = a \diamond b^{-1}.$ $s=\dfrac{80}{\sqrt{13}}-\dfrac{34}{\sqrt{3}}.$ 因此所求答案为 $80+13+34+3=130$。于是 $\begin{aligned} \log_7(a_{2019}) &= \log_7((2019 \heartsuit 2018)\diamond\cdots\diamond(4\heartsuit 3)\diamond(3\heartsuit 2))\\ &= \log_7((2019\diamond 2018^{-1})\diamond\cdots\diamond(4\diamond 3^{-1})\diamond(3\diamond 2^{-1}))\\ &= \log_7(2019\diamond(2018^{-1}\diamond 2018)\diamond\cdots\diamond(3^{-1}\diamond 3)\diamond 2^{-1})\\ &= \log_7(2019\diamond 7\diamond\cdots\diamond 7\diamond 2^{-1})\\ &= \log_7(2019\diamond 2^{-1})\\ &= \log_7(2019^{\frac{1}{\log_7(2)}})\\ &= \dfrac{\log_7(2019)}{\log_7(2)}\\ &= \log_2(2019). \end{aligned}$ 因为 $2^{11}=2048$，所以与 $\log_2(2019)$ 最接近的整数是 $11$。（更准确地说，$\log_2(2019)<\log_2(2048)=11$，并且 $2^{10.5}=2^{10}\cdot\sqrt{2}<1200\cdot 1.5=1800,$ 所以 $\log_2(2019)>\log_2(1800)>10.5。）

Topics

AL.015 Logarithms (rare)