Problem:
Let x,y, and z be positive real numbers that satisfy
2logx​(2y)=2log2x​(4z)=log2x4​(8yz)î€ =0
The value of xy5z can be expressed in the form 2p/q1​, 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
2⋅(ab+1​)=2⋅(a+1c+2​)=4a+1b+c+3​.
Because ab+1​=a+1c+2​, it follows that ab+1​=a+1c+2​=a+a+1b+1+c+2​. Thus 2â‹…(2a+1b+c+3​)=4a+1b+c+3​, and because b+c+3î€ =0, this implies that 8a+2= 2a+1, so a=−61​. Thus −61​b+1​=−61​+1c+2​, which implies that c=−5b−7. Every triple of the form (x,y,z)=(2−1/6,2b,2−5b−7) is a solution to the given system, and hence xy5z=243/61​. Thus p+q=43+6=49​.
The problems on this page are the property of the MAA's American Mathematics Competitions