AMC 10 · 2003 · #24

Grade 11 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

logarithm-propertiesam-gm-inequalityoptimization convert-to-algebraextreme-principle ↑ Prerequisites: logarithm-properties

📏 Medium solution 💡 2 insights

Problem

Two numbers a and b satisfy a ≥ b ＞ 1. Over all such pairs, find the largest value the expression log_a(a/b) + log_b(b/a) can take.

Pick an answer.

(A)

-2

(B)

(C)

(D)

(E)

How to solve

Strategy Introduce a Variable

The expression looks like it depends on two independent numbers, but it does not: it only cares about how a and b compare. Tool #15 (Organize Information in More Ways) makes that visible by splitting each logarithm of a quotient into a difference. Tool #4 (Introduce a Variable) then names the single quantity that survives, t = log_a b, collapsing a two-variable problem into one variable. Once the whole thing is a function of t alone, Tool #14 (Extreme Principle) supplies the ceiling through the classic fact that t + 1/t ≥ 2 for positive t. A maximum claim is only half proved by a ceiling, so Tool #6 (Guess and Check) finishes the job by exhibiting an allowed pair that actually hits the ceiling.

1STEP 1

Split each logarithm of a quotient

Splitting each quotient and using the base rule reduces it to 2 minus a sum.

log_a(a/b) + log_b(b/a) = 2 - (log_a b + log_b a)

2STEP 2

Collapse to one variable

The two logarithms are reciprocals, so one variable t with 0 ＜ t ≤ 1 carries everything.

t = log_a b = (ln b)/(ln a), 0 ＜ t ≤ 1, expression = 2 - (t + 1/t)

3STEP 3

Bound the parenthesis from below

A square being nonnegative shows the sum is at least 2, so the expression is at most 0.

(√(t) - 1/√(t))² ≥ 0 ⟹ t + 1/t ≥ 2 ⟹ 2 - (t + 1/t) ≤ 0

4STEP 4

Show the ceiling is reached

Equality needs the two numbers equal, which is allowed, so 0 is actually reached, choice (B).

a = b = 2: log₂ 1 + log₂ 1 = 0 → (B) 0

Answer

Every choice larger than 0 is impossible, because the expression equals 2 - (t + 1/t) and t + 1/t can never dip below 2 for positive t; so (C), (D), (E) are all out. Choice (A) -2 is a value the expression can take, but it is not the largest: a sample pair shows the expression sitting above it. With a = 4 and b = 2, log₄ 2 + log₂ 1/2 = 1/2 - 1 = -1/2, which already beats -2, and pushing b toward a pushes the value up to 0. Also note the value can be made as negative as we like by taking a enormous compared to b (say a = 2¹⁰⁰, b = 2, giving t = 1/100 and a value near -98), so the expression has no lower bound at all, only the upper bound 0.

💡Key takeaway

When an expression only cares how two numbers compare, name that comparison with one letter — then a number plus its reciprocal is never less than 2, and it hits 2 exactly when the two numbers are equal.

Split each logarithm of a quotient
Collapse to one variable
Bound the parenthesis from below
Show the ceiling is reached

Split each logarithm of a quotient

Collapse to one variable

Bound the parenthesis from below

Show the ceiling is reached

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