Problem:
The system of equations
log10​(2000xy)−(log10​x)(log10​y)log10​(2yz)−(log10​y)(log10​z)log10​(zx)−(log10​z)(log10​x)​=4=1=0​
has two solutions (x1​,y1​,z1​) and (x2​,y2​,z2​). Find y1​+y2​.
Solution:
Let u=log10​x,v=log10​y, and u=log10​z. The given equations can be rewritten as
u−u−v+1u−v−w+1wu−w−u+1​=log10​2=log10​2=1​
and then as
(u−1)(v−1)(u−1)(w−1)(w−1)(u−1)​=log10​2=log10​2=1​
It follows from the first two equations that u=w, and the third equation then implies that u=w=2 or u=u=0. In the first case, v=log10​20, so x1​=100,y1​=20, and z1​=100. In the second case, i=log10​5, so x2​=1,y2​=5, and z2​=1. Thus y1​+y2​=25​.
The problems on this page are the property of the MAA's American Mathematics Competitions