Maths Olympiad Prep

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, 2023

Algebra Difficulty 5.1 AIME, harder Prove it United States

Problem:
Suppose xx, yy, and zz are real numbers greater than 11 such that
xlogyz=2ylogzx=4, and zlogxy=8 \begin{aligned} & x^{\log_{y} z} = 2 \\ & y^{\log_{z} x} = 4, \text{ and } \\ & z^{\log_{x} y} = 8 \end{aligned}
Compute logxy\log_{x} y.

Solution

Solution:
Taking log2\log_{2} both sides of the first equation gives
log2xlogyz=1log2xlog2zlog2y=1 \begin{aligned} & \log_{2} x \log_{y} z = 1 \\ & \frac{\log_{2} x \log_{2} z}{\log_{2} y} = 1 \end{aligned}
Performing similar manipulations on the other two equations, we get
log2xlog2zlog2y=1log2ylog2xlog2z=2log2zlog2ylog2x=3 \begin{aligned} & \frac{\log_{2} x \log_{2} z}{\log_{2} y} = 1 \\ & \frac{\log_{2} y \log_{2} x}{\log_{2} z} = 2 \\ & \frac{\log_{2} z \log_{2} y}{\log_{2} x} = 3 \end{aligned}
Multiplying the first and second equation gives (log2x)2=2\left(\log_{2} x\right)^{2} = 2 or log2x=±2\log_{2} x = \pm \sqrt{2}. Multiplying the second and third equation gives (log2y)2=6\left(\log_{2} y\right)^{2} = 6 or log2y=±6\log_{2} y = \pm \sqrt{6}. Thus, we have
logxy=log2ylog2x=±62=±3 \log_{x} y = \frac{\log_{2} y}{\log_{2} x} = \pm \frac{\sqrt{6}}{\sqrt{2}} = \pm \sqrt{3}

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