AMC 10 · 2015 · #14

Grade 11 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
logarithm-propertiesexponentsdomain-restriction convert-to-algebra ↑ Prerequisites: logarithm-propertiesexponents
📏 Medium solution 💡 3 insights
Problem
Three reciprocals of logarithms with different bases add to one. Find the unknown.

Pick an answer.

(A)
9
(B)
12
(C)
18
(D)
24
(E)
36
How to solve
Strategy Introduce a Variable

The unknown is trapped in the argument slot of three logarithms with three different bases, so there is nothing to combine: log₂ a, log₃ a, log₄ a obey no rule that mixes different bases. The way out is to move a into the base, and the standard move for that is the flip rule 1/(log_b a) = log_a b. That rule is exactly the step this problem turns on, and it is the step most write-ups assert instead of proving — so Tool #4 (Introduce a Variable) names u = log₂ a, which converts the rule into the one-line exponential statement 2^u = a and, in the process, exposes the hidden requirement u ≠ 0 that the rule silently needs. Once the three terms are log_a 2, log_a 3, log_a 4, Tool #15 (Organize Information in More Ways) re-reads the equation: those three numbers are exponents on a, and three exponents adding to 1 is a statement about a product of powers, not a sum of logarithms. Tool #6 (Guess and Check) then puts the candidate back into the original equation, because the derivation only proves that a must be that number, not that it works. Tool #3 (Eliminate Possibilities) closes the loop on the trap: the bases 2, 3, 4 combine by multiplying, and the answer list contains their sum as well as their product.

1STEP 1

Pin down which a are legal

The logarithms fix which values are legal.

a ＞ 0 and a ≠ 1, since log_b 1 = 0 makes 1/(log_b a) undefined
2STEP 2

Prove the flip rule, do not cite it

Each reciprocal is a swapped logarithm.

u = log₂ a ⇔ 2^u = a; u ≠ 0 → a¹/u = 2 ⇔ log_a 2 = 1/(log₂ a)
3STEP 3

Exponents that add mean powers that multiply

Adding exponents means multiplying the arguments.

a^log_a 2 · a^log_a 3 · a^log_a 4 = a^log_a 2 + log_a 3 + log_a 4 = a¹ = a, 2 · 3 · 4 = 24
4STEP 4

Put the value back in

Substituting back confirms the value.

24^S = 2 · 3 · 4 = 24 = 24¹ → S = 1
5STEP 5

Rule out the additive trap

The answer is 24, choice (A).

(ln 24)/(ln 9) ≈ 1.446, (ln 24)/(ln 12) ≈ 1.279, (ln 24)/(ln 18) ≈ 1.100, (ln 24)/(ln 24) = 1, (ln 24)/(ln 36) ≈ 0.887
Answer
24
The cleanest sanity check is that the whole equation, after the flip rule, says log_a 24 = 1 — literally 'raising a to the first power gives 24'. An answer that is not the product 2 · 3 · 4 could not possibly satisfy that, and 24 is on the list. Size also checks out: the left side equals (ln 24)/(ln a), which is greater than 1 for 1 ＜ a ＜ 24 and less than 1 for a ＞ 24, so the solution had to sit strictly inside the answer range between 9 and 36 — and it does. Two honest caveats. First, the flip rule 1/(log_b a) = log_a b is not free: it divides by log_b a, so it is valid only for a ≠ 1. That is harmless here only because the original equation already divides by the same quantity, so a = 1 was never a candidate; a write-up that applies the rule without saying so is asserting a step that can fail. Second, the equation has no solution with 0 ＜ a ＜ 1, since there ln a ＜ 0 makes the left side negative — worth stating, because 'find the value' presumes a uniqueness that the derivation, not the problem statement, supplies.
💡Key takeaway

When the unknown is stuck inside the logarithms, flip each fraction into a logarithm with the unknown as the base — then the three logarithms adding to 1 just means the bases 2, 3, and 4 multiply.

  • Pin down which a are legal
  • Prove the flip rule, do not cite it
  • Exponents that add mean powers that multiply
  • Put the value back in
  • Rule out the additive trap

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