Maths Olympiad Prep

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Algebra Difficulty 4.3 AIME Prove it United States

Problem:
Determine, with proof, the value of log⁡23log⁡34log⁡45…log⁡255256\log_{2} 3 \log_{3} 4 \log_{4} 5 \ldots \log_{255} 256.

Solution

Solution:
We use the fact that log⁡ba=log⁡alog⁡b\log_{b} a = \frac{\log a}{\log b}. Thus, the product equals
log⁡3log⁡2log⁡4log⁡3…log⁡256log⁡255=log⁡256log⁡2=log⁡2(256)=8 \frac{\log 3}{\log 2} \frac{\log 4}{\log 3} \ldots \frac{\log 256}{\log 255} = \frac{\log 256}{\log 2} = \log_{2}(256) = 8

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