2019 AMC 12A Problem 12 — Solution | LIVE by Po-Shen Loh

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2019 AMC 12A Problem 12

Below is the professionally curated solution for Problem 12 of the 2019 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 12A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:logarithm system of equations

Difficulty rating: 1560

12.

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

$\dfrac{25}{2}$

$20$

$\dfrac{45}{2}$

$25$

$32$

Solution:

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

Thus, the correct answer is B.

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