2003 AMC 12A Problem 24

Attempt Problem 24 of the 2003 AMC 12A 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 2003 AMC 12A solutions, or check the answer key.

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24.

If a≥b>1,a \ge b \gt 1, what is the largest possible value of log⁡a(ab)+log⁡b(ba)?\log_a(\frac{a}{b}) + \log_b(\frac{b}{a})?
−2-2
00
22
33
44
Answer: B
Concepts:logarithmAM-GM Inequality
Small hint:
Simplify to 2−log⁡ab−log⁡ba2-\log_a b-\log_b a
Big hint:
With c=log⁡ab>0,c=\log_a b\gt0, use c+1c≥2c+\dfrac1c\ge2
Solution:
Expand: log⁡aab+log⁡bba=(1−log⁡ab)+(1−log⁡ba)=2−(log⁡ab+log⁡ba). \begin{aligned} &\log_a\dfrac ab+\log_b\dfrac ba \\ &\quad {}=(1-\log_a b)+(1-\log_b a) \\ &\quad {}=2-\left(\log_a b+\log_b a\right). \end{aligned}
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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