AMC 10 · 2014 · #18

Grade 11 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

logarithm-propertiesdomain-restrictionexponential-functioninterval-arithmetic work-backwardsidentify-subproblems ↑ Prerequisites: logarithm-propertiesdomain-restriction

📏 Long solution 💡 3 insights

Problem

Five logarithms of different bases are stacked inside one another. Find the length of the interval where the stack is real.

Pick an answer.

(A)

(B)

(C)

271

(D)

319

(E)

511

How to solve

Strategy Work Backwards

Everything known sits on the outside of the expression, and the unknown x sits at the very bottom, so the natural direction of travel is outside-in. The outermost logarithm says only one thing, that its input is positive, and that single fact translates into a condition on the next layer down, which translates again, four times over, until it becomes a condition on x. To keep that bookkeeping readable I first name the five layers, which turns a four-deep nest of parentheses into four one-line subproblems. The part that actually decides whether the argument works is subtler: a chain of one-way implications would prove only that the domain sits inside some interval, and a set can sit inside an interval of length 15/256 while having a much smaller length, or no length at all if it has holes. So before peeling anything I pin down the two logarithm-inequality rules in their reversible, if-and-only-if form, and check that reading them backwards is legitimate. Then each peel is a set equality, and the final interval is the domain exactly. At the end I test the two endpoints and one interior point directly against the original stack, since an outside-in argument is easiest to check by running the machine forwards.

1STEP 1

Name the five layers

Each layer gets its own name.

a=log₁/16x, b=log₁₆a, c=log₁/4b, d=log₄ c, f(x)=log₁/2d

2STEP 2

Make the log rules reversible

The rules must work in both directions.

b ＞ 1: log_b t ＞ k ⇔ t ＞ b^k; 0 ＜ b ＜ 1: log_b t ＞ k ⇔ 0 ＜ t ＜ b^k

3STEP 3

Peel layer one: c exceeds 1

The first peel demands a value above one.

f(x) is real ⇔ d ＞ 0 ⇔ log₄ c ＞ 0 ⇔ c ＞ 4⁰=1

4STEP 4

Peel layer two: b below one-quarter

The next peel flips the inequality.

c ＞ 1 ⇔ log₁/4b ＞ 1 ⇔ 0 ＜ b ＜ (1/4)¹=1/4

5STEP 5

Peel layer three: a between 1 and 2

The third peel gives a clean range.

0 ＜ b ＜ 1/4 ⇔ 16⁰ ＜ a ＜ 16¹/4 ⇔ 1 ＜ a ＜ 2, 16¹/4=(2⁴)¹/4=2

6STEP 6

Peel layer four: reach x

The last peel reaches the variable itself.

1 ＜ a ＜ 2 ⇔ (1/16)² ＜ x ＜ (1/16)¹ ⇔ 1/256 ＜ x ＜ 1/16

7STEP 7

Test the ends and the middle

A midpoint test confirms the interval.

x=1/64: a=3/2, b≈0.1462, c≈1.387, d≈0.236, f≈2.08

8STEP 8

Subtract to get the length

Subtracting gives a length of 15/256.

1/16-1/256=16/256-1/256=15/256

9STEP 9

Reduce the fraction and add

The fraction is already reduced, giving 271.

15=3·5, 256=2⁸, gcd(15,256)=1, m+n=15+256=271

Answer

271

The domain landing far to the left of 1 is exactly what the structure predicts: the innermost base 1/16 is smaller than 1, so forcing log₁/16x to sit between 1 and 2 forces x down between (1/16)² and (1/16)¹. The denominator 256=2⁸ is also unsurprising, since every base in the problem is a power of 2 and the endpoints are powers of 1/16=2⁻⁴. The strongest check is the answer choices themselves. The peel produces exactly four special numbers, 1, 1/4, 1/16 and 1/256, and each of the five choices is the length of an interval whose endpoints are two of them: (A) 19 is 1/4-1/16=3/16, (B) 31 is 1-1/16=15/16, (D) 319 is 1/4-1/256=63/256, (E) 511 is 1-1/256=255/256, and only (C) 271 is the correct pairing 1/16-1/256=15/256. Each wrong choice is the fingerprint of one specific slip: stopping the peel a floor early, or carrying an inner layer's bound down as though it were a bound on x, or missing one of the three reversals. That also means no partial reasoning lands on the right answer by luck, so the endpoints have to be earned separately. Numerically, the interior point x=1/64 passes every floor with room to spare, while x=2⁻¹³, which sits to the left of the interval, gives a=3.25 ＞ 2, then b≈0.425 ＞ 1/4, then c≈0.617 ＜ 1, then d≈-0.348 ＜ 0, and the outermost logarithm fails, precisely as the peel predicts.

💡Key takeaway

Start at the outside, where the only rule is that a logarithm needs a positive input, and push that one demand down a floor at a time, keeping every step reversible so you end up with the domain itself instead of just a box around it.

Name the five layers
Make the log rules reversible
Peel layer one: c exceeds 1
Peel layer two: b below one-quarter
Peel layer three: a between 1 and 2
Peel layer four: reach x
Test the ends and the middle
Subtract to get the length
Reduce the fraction and add