AMC12 2019 A

AMC12 2019 A · Q12

AMC12 2019 A · Q12. It mainly tests Logarithms (rare), Manipulating equations.

Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2 x = \log_y 16$ and $xy = 64$. What is $\left(\log_2 \frac{x}{y}\right)^2$?

正实数$x \neq 1$和$y \neq 1$满足$\log_2 x = \log_y 16$且$xy = 64$。求$\left(\log_2 \frac{x}{y}\right)^2$的值？

(A) $\frac{25}{2}$ $\frac{25}{2}$

(B) 20 20

(D) 25 25

(E) 32 32

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let $a=\log_2 x$ and $b=\log_2 y$. Then \[ \log_y 16 = 4\log_y 2 = \frac{4}{\log_2 y}=\frac{4}{b}, \] so the equation $\log_2 x=\log_y 16$ becomes $ab=4$. Because $xy=2^a\cdot 2^b=64$, it follows that $a+b=6$. Then \[ \left(\log_2\frac{x}{y}\right)^2=(\log_2 x-\log_2 y)^2 =(a-b)^2 =a^2+b^2-2ab =(a+b)^2-4ab =6^2-4\cdot 4 =20. \] Note: Solving the system of equations $\{ab=4,\ a+b=6\}$ gives $\{a,b\}=\{3\pm \sqrt5,\ 3\mp \sqrt5\}$, and $x=2^a$ and $y=2^b$ satisfy the conditions of the problem.

答案（B）：令 $a=\log_2 x$，$b=\log_2 y$。则 \[ \log_y 16=4\log_y 2=\frac{4}{\log_2 y}=\frac{4}{b}, \] 所以方程 $\log_2 x=\log_y 16$ 变为 $ab=4$。由于 $xy=2^a\cdot 2^b=64$，可得 $a+b=6$。于是 \[ \left(\log_2\frac{x}{y}\right)^2=(\log_2 x-\log_2 y)^2 =(a-b)^2 =a^2+b^2-2ab =(a+b)^2-4ab =6^2-4\cdot 4 =20. \] 注：解方程组 $\{ab=4,\ a+b=6\}$ 得 $\{a,b\}=\{3\pm\sqrt5,\ 3\mp\sqrt5\}$，且 $x=2^a$、$y=2^b$ 满足题目条件。

Topics