Question

正实数 $x \ne 1$ 和 $y \ne 1$ 满足 $\log_2 x = \log_y 16$ 且 $xy = 64$ 。 $\left(\log_2 \dfrac{x}{y}\right)^2$ 是多少？

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

$\dfrac{25}{2}$

$20$

$\dfrac{45}{2}$

$25$

$32$

令 $a = \log_2 x$ 且 $b = \log_2 y$ ，则 $\log_y 16 = \dfrac{4}{b}$ ，所以 $a = \dfrac{4}{b}$ ，从而 $ab = 4$ 。

因为 $xy = 64$ ，所以 $a + b = 6$ 。

因此 $\begin{aligned} \left(\log_2 \tfrac{x}{y}\right)^2 &= (a - b)^2 \\ &= (a + b)^2 - 4ab \\ &= 36 - 16 = 20. \end{aligned}$

所以正确答案是 B。

Let $a = \log_2 x$ and $b = \log_2 y.$ Then $\log_y 16 = \dfrac{4}{b},$ so $a = \dfrac{4}{b},$ giving $ab = 4.$

Since $xy = 64,$ we have $a + b = 6.$

Therefore $\begin{aligned} \left(\log_2 \tfrac{x}{y}\right)^2 &= (a - b)^2 \\ &= (a + b)^2 - 4ab \\ &= 36 - 16 = 20. \end{aligned}$

Thus, the correct answer is B.