Problem:
Positive numbers x,y, and z satisfy xyz=1081 and (log10x)(log10yz)+(log10y)(log10z)=468. Find (log10x)2+(log10y)2+(log10z)2.
Solution:
Let a=log10x,b=log10y, and c=log10z. Take the log of each side of the equation xyz=1081 to obtain log10xyz=a+b+c=81. Now square each side of this equation to obtain a2+b2+c2+2ab+2ac+2bc=812. Note that (log10x)(log10yz)=(log10x)(log10y+log10z)=ab+ac. Thus (log10x)2+(log10y)2+(log10z)2=812−2⋅468=5625, and the answer is 5625=75. Note: There are an infinite number of values of a,b, and c which satisfy the conditions of the problem, including a=69,b=6+611, and c=6−611.