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2003 AMC 12A Problem 24

Below is the professionally curated solution for Problem 24 of the 2003 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12A solutions, or check the answer key.

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Concepts:logarithmAM-GM Inequality

Difficulty rating: 2170

24.

If a≥b>1,a \ge b \gt 1, what is the largest possible value of log⁡a(a/b)+log⁡b(b/a)?\log_a(a/b) + \log_b(b/a)?

−2-2

00

22

33

44

Solution:

Expand: log⁡aab+log⁡bba=(1−log⁡ab)+(1−log⁡ba)=2−(log⁡ab+log⁡ba).\log_a\dfrac ab+\log_b\dfrac ba=(1-\log_a b)+(1-\log_b a)=2-\left(\log_a b+\log_b a\right).

Let c=log⁡ab>0.c=\log_a b\gt0. Since c+1c≥2c+\dfrac1c\ge2 by AM-GM, the expression is at most 0.0.

Equality holds when c=1,c=1, that is, when a=b,a=b, so the largest value is 0.0.

Thus, the correct answer is B.

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