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Algebra Difficulty 4.8 AIME Prove it Ukraine

Compare the following numbers: A=11A = 11, B=log⁡23⋅log⁡34⋅log⁡45⋯log⁡20152016B = \log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{2015} 2016 and C=log⁡32⋅log⁡43⋅log⁡54⋯log⁡20162015C = \log_3 2 \cdot \log_4 3 \cdot \log_5 4 \cdots \log_{2016} 2015.

Solution

Obviously, for any integer n>0n > 0 log⁡n+1n<1\log_{n+1} n < 1. Hence, C<1C < 1. Also we have that B⋅C=1B \cdot C = 1, so B>1B > 1, moreover
B=lg⁡3lg⁡2⋅lg⁡4lg⁡3⋅lg⁡5lg⁡4⋯lg⁡2016lg⁡2015=lg⁡2016lg⁡2<11=A⇔lg⁡2016<11lg⁡2=lg⁡211⇔2016<211=2048. B = \frac{\lg 3}{\lg 2} \cdot \frac{\lg 4}{\lg 3} \cdot \frac{\lg 5}{\lg 4} \cdots \frac{\lg 2016}{\lg 2015} = \frac{\lg 2016}{\lg 2} < 11 = A \Leftrightarrow \lg 2016 < 11 \lg 2 = \lg 2^{11} \Leftrightarrow 2016 < 2^{11} = 2048.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.