Problem:
Let x,y, and z be positive real numbers that satisfy
2logxβ(2y)=2log2xβ(4z)=log2x4β(8yz)ξ =0
The value of xy5z can be expressed in the form 2p/q1β, where p and q are relatively prime positive integers. Find p+q.
Solution:
Let x=2a,y=2b, and z=2c. Then the given equality becomes
2β
(ab+1β)=2β
(a+1c+2β)=4a+1b+c+3β.
Because ab+1β=a+1c+2β, it follows that ab+1β=a+1c+2β=a+a+1b+1+c+2β. Thus 2β
(2a+1b+c+3β)=4a+1b+c+3β, and because b+c+3ξ =0, this implies that 8a+2= 2a+1, so a=β61β. Thus β61βb+1β=β61β+1c+2β, which implies that c=β5bβ7. Every triple of the form (x,y,z)=(2β1/6,2b,2β5bβ7) is a solution to the given system, and hence xy5z=243/61β. Thus p+q=43+6=49β.
The problems on this page are the property of the MAA's American Mathematics Competitions