2018 AIME I Problems/Problem 5

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 $K$ .

Solution 1

Using the logarithmic property $\log_{a^n}b^n = \log_{a}b$ , we note that $(2x+y)^2 = 4x^2+4xy+y^2$ . 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 $\boxed{189}$ .

-expiLnCalc

Note

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

Solution 2

Do as done in Solution 1 to get $x^2+xy-2y^2=0\implies (\frac{x}{y})^2+\frac{x}{y}-2=0\implies \frac{x}{y}=\frac{-1\pm\sqrt{1+8}}{2}=1,-2$ . Do as done in Solution 1 to get $9x^2+6xy+y^2=3x^2+4xy+Ky^2\implies 6x^2+2xy+(1-K)y^2=0\implies 6(\frac{x}{y})^2+2\frac{x}{y}+(1-K)=0\implies \frac{x}{y}=\frac{-2\pm \sqrt{4-24(1-K)}}{12}\implies \frac{x}{y}=\frac{-2\pm 2\sqrt{6K-5}}{12}=\frac{-1\pm \sqrt{6K-5}}{6}$ . If $\frac{x}{y}=1$ , then $1=\frac{-1\pm \sqrt{6K-5}}{6}\implies 6=-1\pm \sqrt{6K-5}\implies 7=\pm \sqrt{6K-5}\implies 49=6K-5\implies K=9$ . If $\frac{x}{y}=-2$ , then $-2=\frac{-1\pm \sqrt{6K-5}}{6}\implies -12=-1\pm \sqrt{6K-5}\implies -11=\sqrt{6K-5}\implies 121=6K-5\implies 126=6K\implies K=21$ . Hence our final answer is $21\cdot 6=\boxed{126}$ -vsamc

2018 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2018 AIME I Problems/Problem 5

Contents

Problem 5

Solution 1

Note

Solution 2

See Also