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Harvard-MIT Mathematics Tournament

Algebra Difficulty 5.5 AIME, harder Prove it United States

Problem:

Let $\otimes$ be a binary operation that takes two positive real numbers and returns a positive real number. Suppose further that $\otimes$ is continuous, commutative $(a \otimes b = b \otimes a)$ , distributive across multiplication $(a \otimes (b c) = (a \otimes b)(a \otimes c))$ , and that $2 \otimes 2 = 4$ . Solve the equation $x \otimes y = x$ for $y$ in terms of $x$ for $x > 1$ .

Solution

Solution:

Answer: $y = \sqrt{2}$

We note that $\left(a \otimes b^{k}\right) = (a \otimes b)^{k}$ for all positive integers $k$ . Then for all rational numbers $\frac{p}{q}$ we have $a \otimes b^{\frac{p}{q}} = \left(a \otimes b^{\frac{1}{q}}\right)^{p} = (a \otimes b)^{\frac{p}{q}}$ . So by continuity, for all real numbers $a, b$ , it follows that $2^{a} \otimes 2^{b} = (2 \otimes 2)^{a b} = 4^{a b}$ . Therefore given positive reals $x, y$ , we have

$x \otimes y = 2^{\log_{2}(x)} \otimes 2^{\log_{2}(y)} = 4^{\log_{2}(x) \log_{2}(y)}$ .

If $x = 4^{\log_{2}(x) \log_{2}(y)} = 2^{2 \log_{2}(x) \log_{2}(y)}$ then $\log_{2}(x) = 2 \log_{2}(x) \log_{2}(y)$ and $1 = 2 \log_{2}(y) = \log_{2}\left(y^{2}\right)$ . Thus $y = \sqrt{2}$ regardless of $x$ .

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