2018 AIME I Problems/Problem 5

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Problem 5

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 $K$.


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 $\boxed{189}$.



The cases $x=y$ and $x=-2y$ can be found by SFFT (Simon's Favorite Factoring Trick) from $x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0$.

See Also

2018 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AIME Problems and Solutions

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