SpecialEquation LogAndExp AIME Intermediate
2018


Problem - 4105

For each ordered pair of real numbers $(x,y)$ satisfying\[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\]there is a real number $K$ such that\[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\]Find the product of all possible values of $K$.


Answer     189

Note that $(2x+y)^2 = x^2+xy+7y^2$. That gives $x^2+xy-2y^2=0$ upon simplification and division by $3$. Then, $x=y$ or $x=-2y$. From the second equation, $9x^2+6xy+y^2=3x^2+4xy+Ky^2$. If we take $x=y$, we see that $K=9$. If we take $x=-2y$, we see that $K=21$. The product is $\boxed{189}$.

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