2008 AMC 12A Problem 16

Attempt Problem 16 of the 2008 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12A solutions, or check the answer key.

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16.

The numbers log⁡(a3b7),\log(a^3 b^7), log⁡(a5b12),\log(a^5 b^{12}), and log⁡(a8b15)\log(a^8 b^{15}) are the first three terms of an arithmetic sequence, and the 1212th term of the sequence is log⁡(bn).\log(b^n). What is n?n?
4040
5656
7676
112112
143143
Answer: D
Concepts:arithmetic sequencelogarithm
Small hint:
Write each term as plog⁡a+qlog⁡bp \log a + q \log b and set the consecutive differences equal
Big hint:
Equal differences force log⁡a=2log⁡b,\log a = 2 \log b, reducing every term to a multiple of log⁡b\log b
Solution:
The three terms are 3log⁡a+7log⁡b,3\log a + 7\log b, 5log⁡a+12log⁡b,5\log a + 12\log b, and 8log⁡a+15log⁡b.8\log a + 15\log b. Setting the two consecutive differences equal, 2log⁡a+5log⁡b=3log⁡a+3log⁡b, \begin{aligned} &2\log a + 5\log b \\ &= 3\log a + 3\log b, \end{aligned} so log⁡a=2log⁡b.\log a = 2\log b.
The first term is then (3⋅2+7)log⁡b=13log⁡b,(3 \cdot 2 + 7)\log b = 13\log b, and the common difference is (2⋅2+5)log⁡b=9log⁡b.(2 \cdot 2 + 5)\log b = 9\log b.
The 1212th term is (13+11⋅9)log⁡b=112log⁡b=log⁡(b112), \begin{aligned} (13 + 11 \cdot 9)\log b &= 112\log b \\ &= \log(b^{112}), \end{aligned} so n=112.n = 112.
Thus, D is the correct answer.
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