Problem:
Let x,y, and z be positive real numbers that satisfy the following system of equations:
log2(yzx)log2(xzy)log2(xyz)=21=31=41.
Then the value of ∣∣∣log2(x4y3z2)∣∣∣ is nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let a=log2x,b=log2y, and c=log2z. Using the rules for the logarithm of a product and quotient gives the following system of equations:
a−b−cb−a−cc−a−b=21=31=41
Solving this system yields a=−247,b=−83, and c=−125. Thus
∣log2(x4y3z2)∣=∣4a+3b+2c∣=∣∣∣∣∣−67−89−65∣∣∣∣∣=∣∣∣∣∣−825∣∣∣∣∣=825.
The requested sum is 25+8=33.
The problems on this page are the property of the MAA's American Mathematics Competitions