Problem:
The system of equations
log10(2000xy)−(log10x)(log10y)log10(2yz)−(log10y)(log10z)log10(zx)−(log10z)(log10x)=4=1=0
has two solutions (x1,y1,z1) and (x2,y2,z2). Find y1+y2.
Solution:
Let u=log10x,v=log10y, and u=log10z. The given equations can be rewritten as
u−u−v+1u−v−w+1wu−w−u+1=log102=log102=1
and then as
(u−1)(v−1)(u−1)(w−1)(w−1)(u−1)=log102=log102=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=log1020, so x1=100,y1=20, and z1=100. In the second case, i=log105, 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