Problem:
For each ordered pair of real numbers (x,y) satisfying
log2β(2x+y)=log4β(x2+xy+7y2),
there is a real number K such that
log3β(3x+y)=log9β(3x2+4xy+Ky2).
Find the product of all possible values of K.
Solution:
Because x2+xy+7y2=(x+2yβ)2+427βy2>0, the right side of the first equation is real. It follows that the left side of the equation is also real, so 2x+y>0 and
log2β(2x+y)=log22β(2x+y)2=log4β(4x2+4xy+y2).
Thus 4x2+4xy+y2=x2+xy+7y2, which implies that 0=x2+xyβ2y2= (x+2y)(xβy). Therefore either x=β2y or x=y, and because 2x+y>0, x must be positive and 3x+y=x+(2x+y)>0. Similarly,
log3β(3x+y)=log32β(3x+y)2=log9β(9x2+6xy+y2).
If x=β2yξ =0, then 9x2+6xy+y2=36y2β12y2+y2=25y2= 3x2+4xy+Ky2 when K=21. If x=yξ =0, then 9x2+6xy+y2= 16y2=3x2+4xy+Ky2 when K=9. The requested product is 21β
9=189β.
The problems on this page are the property of the MAA's American Mathematics Competitions