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 $K$ .

Solutions

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 $\boxed{189}$ .

2018 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2018 AIME I Problems/Problem 5

Solutions

Straightforward Solution

See Also