AMC 10 · 2004 · #16

Grade 11 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

logarithm-propertiesdomain-restriction work-backwardseasier-related-problem ↑ Prerequisites: exponents

📏 Medium solution 💡 2 insights

Problem

The expression log₂₀₀₄(log₂₀₀₃(log₂₀₀₂(log₂₀₀₁{x}))) has four logarithms nested inside one another. It makes sense only for large enough x, and the set of x that work is an open ray. Find the threshold c.

Pick an answer.

(A)

(B)

$2001^{2002}$

(C)

$2002^{2003}$

(D)

$2003^{2004}$

(E)

$2001^{2002^{2003}}$

How to solve

Strategy Work Backwards

Nothing here can be attacked from the x end: you cannot tell whether a particular x works without evaluating all four layers. But the outermost layer states its requirement immediately — whatever sits inside it must be positive — and that requirement is an inequality about the next layer in. So the natural direction is backwards (Tool #11): start at the outside, turn each layer's requirement into a requirement on the layer beneath it, and keep going until the requirement is about x itself. Tool #7 (Identify Subproblems) makes this bookkeeping safe by naming the four layers L₁,…,L₄ so no bracket gets lost. Tool #9 (Solve an Easier Related Problem) supplies the pattern: work out the one-layer and two-layer versions first, where the answer is obvious, and the four-layer version becomes the same move repeated. The one thing to watch is that each move must be an equivalence, not just an implication, or the ray you end up with could be too big or too small.

1STEP 1

Name the four layers

Name the four layers from the inside out so no bracket is lost.

L₁=log₂₀₀₁x, L₂=log₂₀₀₂L₁, L₃=log₂₀₀₃L₂, L₄=log₂₀₀₄L₃

2STEP 2

Try one and two layers first

Small cases reveal the move: a demand on a layer becomes a power of the base on its input.

log₂₀₀₁x defined ⇔ x ＞ 0; log₂₀₀₂(log₂₀₀₁x) defined ⇔ L₁ ＞ 0 ⇔ x ＞ 2001⁰=1

3STEP 3

Peel the outermost layer

Peeling the outermost layer requires the next one to exceed 1.

log₂₀₀₃L₂ ＞ 0 ⇔ L₂ ＞ 2003⁰=1

4STEP 4

Peel the next layer

The next peel jumps the threshold to 2002.

log₂₀₀₂L₁ ＞ 1 ⇔ L₁ ＞ 2002¹=2002

5STEP 5

Peel the last layer to reach x

One final peel reaches x, giving 2001²⁰⁰².

log₂₀₀₁x ＞ 2002 ⇔ x ＞ 2001²⁰⁰²

6STEP 6

Check the chain runs both ways, then read off c

Every step was an equivalence, so that is exactly the threshold, choice (B).

log₂₀₀₄(log₂₀₀₃(log₂₀₀₂(log₂₀₀₁x))) defined⇔ x ＞ 2001²⁰⁰² → c=2001²⁰⁰² (B)

Answer

2001²⁰⁰²

Test the boundary. At x=2001²⁰⁰² the layers are L₁=2002, L₂=log₂₀₀₂₂₀₀₂=1, L₃=log₂₀₀₃₁=0, and then L₄=log₂₀₀₄₀ is undefined — so x=2001²⁰⁰² genuinely fails, matching a strict inequality x ＞ c rather than x ≥ c. Just above it, say x=2001²⁰⁰³, we get L₁=2003, L₂=log₂₀₀₂₂₀₀₃ slightly more than 1, L₃ a small positive number, and L₄ a large negative number that is perfectly defined. So the threshold is exactly right and not off by a layer. The wrong choices are each a specific slip: (A) 0 is the answer if only the innermost logarithm is considered; (C) 2002²⁰⁰³ and (D) 2003²⁰⁰⁴ pair the wrong base with the wrong exponent; and (E) 2001^{2002²⁰⁰³} is what you get by peeling one layer too many, treating the outermost logarithm as if it also had to exceed 1 instead of merely being positive.

💡Key takeaway

Peel a stack of logarithms from the outside in: each layer only demands that the thing inside it be positive, and one layer down that demand becomes "bigger than the base raised to that power".

Name the four layers
Try one and two layers first
Peel the outermost layer
Peel the next layer
Peel the last layer to reach x
Check the chain runs both ways, then read off c