AMC12 2022 B

AMC12 2022 B · Q16

AMC12 2022 B · Q16. It mainly tests Quadratic equations, Logarithms (rare).

Suppose $x$ and $y$ are positive real numbers such that \[x^y=2^{64}\text{ and }(\log_2{x})^{\log_2{y}}=2^{7}.\] What is the greatest possible value of $\log_2{y}$?

假设 $x$ 和 $y$ 是正实数，使得 \[x^y=2^{64}\text{ 且 }(\log_2{x})^{\log_2{y}}=2^{7}.\] $\log_2{y}$ 的最大可能值为多少？

(A) 3 3

(B) 4 4

(D) 4+\sqrt{3} 4+\sqrt{3}

(E) 7 7

Answer

Correct choice: (C)

正确答案：（C）

Solution

Take the base-two logarithm of both equations to get \[y\log_2 x = 64\quad\text{and}\quad (\log_2 y)(\log_2\log_2 x) = 7.\] Now taking the base-two logarithm of the first equation again yields \[\log_2 y + \log_2\log_2 x = 6.\] It follows that the real numbers $r:=\log_2 y$ and $s:=\log_2\log_2 x$ satisfy $r+s=6$ and $rs = 7$. Solving this system yields \[\{\log_2 y,\log_2\log_2 x\}\in\{3-\sqrt 2, 3 + \sqrt 2\}.\] Thus the largest possible value of $\log_2 y$ is $3+\sqrt 2 \implies \boxed{\textbf{(C) }3+\sqrt{2}}$.

对两个方程取以 2 为底的对数，得到 \[y\log_2 x = 64\quad\text{且}\quad (\log_2 y)(\log_2\log_2 x) = 7.\] 再次对第一个方程取以 2 为底的对数，得到 \[\log_2 y + \log_2\log_2 x = 6.\] 由此可知实数 $r:=\log_2 y$ 和 $s:=\log_2\log_2 x$ 满足 $r+s=6$ 和 $rs = 7$。解此方程组得 \[\{\log_2 y,\log_2\log_2 x\}\in\{3-\sqrt 2, 3 + \sqrt 2\}.\] 因此 $\log_2 y$ 的最大可能值为 $3+\sqrt 2 \implies \boxed{\textbf{(C) }3+\sqrt{2}}$。

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