Problem:
Let x,y, and z be real numbers satisfying the system
​log2​(xyz−3+log5​x)=5log3​(xyz−3+log5​y)=4log4​(xyz−3+log5​z)=4​
Find the value of ∣log5​x∣+∣log5​y∣+∣log5​z∣.
Solution:
The system is equivalent to
​xyz+log5​x=35xyz+log5​y=84xyz+log5​z=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