Positive real numbers x ≠ 1 and y ≠ 1 satisfy \log_2 x = \log_y 16 and xy = 64. What is (\log_2 (x)/(y))^2?

2019 AMC 12A Problem 12 — Solution

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12.

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

$\dfrac{25}{2}$

$20$

$\dfrac{45}{2}$

$25$

$32$

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.