AMC 10 · 2018 · #7

Grade 11 algebra

Archetype Telescoping Collapse └ Sub-archetype Product Ratio Cancellation

logarithm-propertiestelescoping-sumsequences-arithmeticpattern-recognition pattern-recognitionidentify-subproblems ↑ Prerequisites: logarithm-propertiessequences-arithmetic

📏 Medium solution 💡 3 insights

Problem

Eleven logarithms are multiplied together. Their bases run through the odd numbers 3, 5, 7, and so on up to 23, and each factor's argument is four more than its own base: base three of seven, then base five of nine, then base seven of eleven, and so on up to base twenty-three of twenty-seven. Find the value of that product.

Pick an answer.

(A)

(B)

$3\log_{7}23$

(C)

(D)

(E)

How to solve

Strategy Organize Information in More Ways

Each factor sits in its own base, and logarithms in different bases have no way to interact. That is the whole difficulty: eleven ugly irrational numbers with nothing joining them. The fix is not to compute them but to re-express them. The Change of Base Formula rewrites every factor as a ratio of logarithms taken in one single shared base, and once all eleven live in the same base the product becomes one tall fraction. Its numerator list and denominator list turn out to be the same run of odd numbers offset by two positions, so almost everything pairs off and cancels. Only the two ends survive, and those two ends are exact powers of each other.

1STEP 1

Write out every hidden factor

Write out every hidden factor.

log₃ 7 · log₅ 9 · log₇ 11 · log₉ 13 · log₁₁ 15 · log₁₃ 17 · log₁₅ 19 · log₁₇ 21 · log₁₉ 23 · log₂₁ 25 · log₂₃ 27

2STEP 2

Move every factor to one base

Move them all to one base.

log_b a = (log a)/(log b) ⟹ log₃ 7 · log₅ 9 … log₂₃ 27 = (log 7)/(log 3) · (log 9)/(log 5) · (log 11)/(log 7) … (log 25)/(log 21) · (log 27)/(log 23)

3STEP 3

Collapse into one big fraction

Collapse into one big fraction.

(log 7 · log 9 · log 11 … log 21 · log 23 · log 25 · log 27)/(log 3 · log 5 · log 7 · log 9 … log 19 · log 21 · log 23)

4STEP 4

Cancel the block both lists share

The shared middle block cancels entirely.

((log 7 · log 9 … log 23) · log 25 · log 27)/(log 3 · log 5 · (log 7 · log 9 … log 23)) = (log 25 · log 27)/(log 3 · log 5)

5STEP 5

Evaluate the four survivors

Only four survivors remain.

(log 25 · log 27)/(log 3 · log 5) = ((2log 5)(3log 3))/(log 3 · log 5) = 2 · 3 = 6

6STEP 6

Match the choices, dodge the traps

Evaluating gives 6.

(log 27)/(log 3) · (log 25)/(log 5) = 3 · 2 = 6 ⟹ (C)

Answer

Two independent checks agree. First, the structure predicts a whole number: the only pieces that escape the cancelling are log 25 and log 27 over log 3 and log 5, and 25 and 27 are exact powers of 5 and 3, so what is left had to come out as 2 · 3. Second, the decimals confirm it. Every factor exceeds 1, since each argument b + 4 is larger than its base b, and the factors shrink steadily toward 1 as the numbers grow: log₃ 7 ≈ 1.771, log₅ 9 ≈ 1.365, log₇ 11 ≈ 1.232, down to log₂₃ 27 ≈ 1.051. Multiplying all eleven gives 5.999…, which is 6 up to rounding. A size check also fits: the product is bigger than 1 but nowhere near 10, which is consistent with 6 and awkward for choice (E). And choice (B) is the only choice that is not a plain number, which is itself a hint that it belongs to an incomplete cancellation rather than to a finished one.

💡Key takeaway

When a long chain of logarithms is multiplied together, rewrite them all in one common base — the middle terms pair off and vanish, and only the two ends decide the answer.

Write out every hidden factor
Move every factor to one base
Collapse into one big fraction
Cancel the block both lists share
Evaluate the four survivors
Match the choices, dodge the traps