Problem:
For each ordered pair of real numbers (x,y) satisfying
log2​(2x+y)=log4​(x2+xy+7y2),
there is a real number K such that
log3​(3x+y)=log9​(3x2+4xy+Ky2).
Find the product of all possible values of K.
Solution:
Because x2+xy+7y2=(x+2y​)2+427​y2>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
log2​(2x+y)=log22​(2x+y)2=log4​(4x2+4xy+y2).
Thus 4x2+4xy+y2=x2+xy+7y2, which implies that 0=x2+xy−2y2= (x+2y)(x−y). Therefore either x=−2y or x=y, and because 2x+y>0, x must be positive and 3x+y=x+(2x+y)>0. Similarly,
log3​(3x+y)=log32​(3x+y)2=log9​(9x2+6xy+y2).
If x=−2yî€ =0, then 9x2+6xy+y2=36y2−12y2+y2=25y2= 3x2+4xy+Ky2 when K=21. If x=yî€ =0, then 9x2+6xy+y2= 16y2=3x2+4xy+Ky2 when K=9. The requested product is 21â‹…9=189​.
The problems on this page are the property of the MAA's American Mathematics Competitions