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2008 AMC 12A Problem 16

Problem 16 of 25IntermediateAlgebra

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?

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Solution

The three terms are 3loga+7logb,3\log a + 7\log b, 5loga+12logb,5\log a + 12\log b, and 8loga+15logb.8\log a + 15\log b. Setting the two consecutive differences equal, 2loga+5logb=3loga+3logb, \begin{aligned} &2\log a + 5\log b \\ &= 3\log a + 3\log b, \end{aligned} so loga=2logb.\log a = 2\log b. The first term is then (32+7)logb=13logb,(3 \cdot 2 + 7)\log b = 13\log b, and the common difference is (22+5)logb=9logb.(2 \cdot 2 + 5)\log b = 9\log b. The 1212th term is (13+119)logb=112logb=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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Concepts: arithmetic sequence · logarithm

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