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Algebra Difficulty 4.9 AIME Prove it Ukraine

Which number is larger, A=19:120233A = \frac{1}{9} : \sqrt[3]{\frac{1}{2023}} or B=log⁡202391125B = \log_{2023} 91125?

Solution

To prove this, we will show that the following inequalities hold: A<32<BA < \frac{3}{2} < B.

A=19:120233=202339<139<32,B=log⁡202391125>log⁡2023453=log⁡452453=32. A = \frac{1}{9} : \sqrt[3]{\frac{1}{2023}} = \frac{\sqrt[3]{2023}}{9} < \frac{13}{9} < \frac{3}{2}, \quad B = \log_{2023} 91125 > \log_{2023} 45^3 = \log_{45^2} 45^3 = \frac{3}{2}.

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