2022 AMC 12B Problem 16

Attempt Problem 16 of the 2022 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12B solutions, or check the answer key.

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16.

Suppose xx and yy are positive real numbers such that

xy=264x^y = 2^{64} and (log2x)log2y=27.(\log_2 x)^{\log_2 y} = 2^7.

What is the greatest possible value of log2y?\log_2 y?

33

44

3+23 + \sqrt2

4+34 + \sqrt3

77

Answer: C
Concepts:logarithmsubstitutionquadratic
Difficulty rating: 1800
Solution:

Let a=log2xa = \log_2 x and b=log2y.b = \log_2 y. Taking log2\log_2 of xy=264x^y = 2^{64} gives ylog2x=64,y \log_2 x = 64, i.e. a2b=26.a \cdot 2^b = 2^6.

Taking log2\log_2 of the second equation gives blog2a=7,b \log_2 a = 7, so a=27/b.a = 2^{7/b}. Substituting, 27/b2b=26,2^{7/b} \cdot 2^b = 2^6, so b+7b=6,b + \dfrac7b = 6, i.e. b26b+7=0.b^2 - 6b + 7 = 0.

Thus b=3±2,b = 3 \pm \sqrt2, and the greatest value of log2y\log_2 y is 3+2.3 + \sqrt2.

Thus, the correct answer is C.

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