AMC 10 · 2008 · #16

Grade 11 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

logarithm-propertiessequences-arithmeticsystems-of-equations convert-to-algebrapattern-recognition ↑ Prerequisites: logarithm-properties

📏 Long solution 💡 2 insights

Problem

Three logarithms of products of two bases are the first three terms of an arithmetic sequence. The twelfth term equals a logarithm of one base to some power. Find that power.

Pick an answer.

(A)

(B)

(C)

(D)

112

(E)

143

How to solve

Strategy Introduce a Variable

The logarithms look complicated, but log(a^x b^y) = x log a + y log b, so every term in the problem is a fixed combination of just two numbers: log a and log b. Naming those two numbers turns the whole problem into linear algebra with two unknowns. The arithmetic condition then becomes a single equation, and once that equation pins down how log a compares to log b, every term collapses into a plain multiple of log b, where counting the pattern is easy. The last move, cancelling log b, is the one place the problem can quietly fail, so it gets checked instead of assumed.

1STEP 1

Name the two logarithms

Naming the two logarithms turns every term into a weighted sum.

log(a^x b^y) = xA + yB, A=log a, B=log b

2STEP 2

Turn arithmetic into one equation

Equal gaps give one equation between them.

(5A+12B)-(3A+7B) = 2A+5B, (8A+15B)-(5A+12B) = 3A+3B

3STEP 3

Solve for how A compares to B

Solving relates the two, so one logarithm carries everything.

2A+5B = 3A+3B → A = 2B ⟺ a = b²

4STEP 4

Rewrite every term as a weight

Every term becomes a plain weight times that logarithm.

log(a^x b^y) = (2x+y)B: 13B, 22B, 31B,… (gap 9B)

5STEP 5

Walk out to the 12th weight

Walking to the twelfth weight gives 112.

w_k = 13 + 9(k-1) = 9k+4, w₁₂ = 13 + 9 · 11 = 112

6STEP 6

Cancel log b honestly

Cancelling honestly confirms 112, choice (A).

nB = 112B → (n-112)B = 0; B ≠ 0 → n = 112

Answer

112

Test a concrete pair satisfying a = b²: take b = 10 and a = 100. Then log(a³ b⁷) = log(10⁶ * 10⁷) = 13, log(a⁵ b¹2) = log(10¹0 * 10¹2) = 22, and log(a⁸ b¹5) = log(10¹6 * 10¹5) = 31. These are 13, 22, 31, arithmetic with difference 9, so the hypothesis really is satisfiable. The 12th term is 13 + 11(9) = 112 = log(10¹12) = log(b¹12), giving n = 112, matching (D). The distractors confirm the shape of the trap rather than the algebra: the weights follow 9k + 4, and 40 = 9(4) + 4 is the 4th term while 76 = 9(8) + 4 is the 8th term, so choices (A) and (C) are the same sequence read at the wrong index. That means a mod-9 check cannot separate them; the only thing that decides the answer is counting eleven gaps from the first term, which is what step 5 does.

💡Key takeaway

Name log a and log b, and a scary sequence of logarithms turns into ordinary counting by 9s.

Name the two logarithms
Turn arithmetic into one equation
Solve for how A compares to B
Rewrite every term as a weight
Walk out to the 12th weight
Cancel log b honestly