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2019 AMC 12A Problem 12

Problem 12 of 25IntermediateAlgebra

Positive real numbers x≠1x \ne 1 and y≠1y \ne 1 satisfy log⁡2x=log⁡y16\log_2 x = \log_y 16 and xy=64.xy = 64. What is (log⁡2xy)2?\left(\log_2 \dfrac{x}{y}\right)^2?

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Solution

Let a=log⁡2xa = \log_2 x and b=log⁡2y.b = \log_2 y. Then log⁡y16=4b,\log_y 16 = \dfrac{4}{b}, so a=4b,a = \dfrac{4}{b}, giving ab=4.ab = 4. Since xy=64,xy = 64, we have a+b=6.a + b = 6. Therefore (log⁡2xy)2=(a−b)2=(a+b)2−4ab=36−16=20. \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.

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Concepts: logarithm · system of equations

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.