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 $a \ge b \gt 1,$ what is the largest possible value of $\log_a(a/b) + \log_b(b/a)?$

$-2$

$0$

$2$

$3$

$4$

Answer: B

Concepts:logarithm AM-GM Inequality

Difficulty rating: 2170

Solution:

Expand: $\begin{aligned} &\log_a\dfrac ab+\log_b\dfrac ba \\ &\quad {}=(1-\log_a b)+(1-\log_b a) \\ &\quad {}=2-\left(\log_a b+\log_b a\right). \end{aligned}$

Let $c=\log_a b\gt0.$ Since $c+\dfrac1c\ge2$ by AM-GM, the expression is at most $0.$

Equality holds when $c=1,$ that is, when $a=b,$ so the largest value is $0.$

Thus, the correct answer is B.

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