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

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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}

You get

ab = 4

and

a + b = 6;

note

\left(\log_2\tfrac{x}{y}\right)^2 = (a - b)^2

Let

a = \log_2 x

and

b = \log_2 y.

Then

\log_y 16 = \dfrac{4}{b},

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.