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2019 AMC 12A Problem 23

Problem 23 of 25HarderAlgebra

Define binary operations ♢\diamondsuit and ♡\heartsuit by a♢b=alog⁡7(b) a \diamondsuit b = a^{\log_7(b)} and a♡b=a1log⁡7(b) a \heartsuit b = a^{\frac{1}{\log_7(b)}} for all real numbers aa and bb for which these expressions are defined. The sequence (an)(a_n) is defined recursively by a3=3♡2a_3 = 3 \heartsuit 2 and an=(n♡(n−1))♢an−1 a_n = (n \heartsuit (n - 1)) \diamondsuit a_{n-1} for all integers n≥4.n \ge 4. To the nearest integer, what is log⁡7(a2019)?\log_7(a_{2019})?

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Solution

Let L(x)=log⁡7x.L(x) = \log_7 x. Then L(a♢b)=L(a)L(b)L(a \diamondsuit b) = L(a)L(b) and L(a♡b)=L(a)L(b).L(a \heartsuit b) = \dfrac{L(a)}{L(b)}. So L(a3)=L(3)L(2),L(a_3) = \dfrac{L(3)}{L(2)}, and L(an)=L(n)L(n−1)⋅L(an−1).L(a_n) = \dfrac{L(n)}{L(n-1)} \cdot L(a_{n-1}). The product telescopes: L(aN)=L(3)L(2)⋅L(N)L(3)=L(N)L(2). \begin{aligned} L(a_N) &= \dfrac{L(3)}{L(2)} \cdot \dfrac{L(N)}{L(3)} \\ &= \dfrac{L(N)}{L(2)}. \end{aligned} Hence L(a2019)=log⁡72019log⁡72L(a_{2019}) = \dfrac{\log_7 2019}{\log_7 2} =log⁡22019≈10.98,= \log_2 2019 \approx 10.98, which rounds to 11.11. Thus, the correct answer is D.
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Tagged: custom operation · logarithm · telescoping

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