Difference between revisions of "2018 AIME I Problems/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 $K$.

Straightforward Solution

Note that $(2x+y)^2 = x^2+xy+7y^2$. DO NOT SQUARE THE WRONG SIDE! 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 $\boxed{189}$.

-expiLnCalc