HMMT-Feb 2012 alg, Problem 7

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: 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 $x$. $\fbox{y=\sqrt{2}}$.

Answer: y=\sqrt{2}