Problem:
Let x,y, and z be real numbers satisfying the system
log2(xyz−3+log5x)=5log3(xyz−3+log5y)=4log4(xyz−3+log5z)=4
Find the value of ∣log5x∣+∣log5y∣+∣log5z∣.
Solution:
The system is equivalent to
xyz+log5x=35xyz+log5y=84xyz+log5z=259.
Let x=5α,y=5β, and z=5γ. Then
5α+β+γ+α=355α+β+γ+β=845α+β+γ+γ=259
Adding these equations yields 3⋅5α+β+γ+(α+β+γ)=378. Let t be chosen so that 3t=α+β+γ. Then 3⋅53t+3t=378 and 125t+t=126. Because 125t+t is an increasing function of t, there is only one value of t satisfying this equation, and inspection shows that t=1. Thus α+β+γ=3, implying that 53+α=35 and α=−90;53+β=84 and β=−41; and 53+γ=259 and γ=134. The requested sum is ∣α∣+∣β∣+∣γ∣=90+41+134=265.
The problems on this page are the property of the MAA's American Mathematics Competitions