HMMT February

Algebra Difficulty 5.2 AIME, harder Find the answer

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 $x>1$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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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