Problem: If β£xβlogyβ£=x+logy where x and logy are real, then
Answer Choices:
A. x=0
B. y=1
C. x=0 and y=1
D. x(yβ1)=0
E. None of these
Solution:
if (xβlogy) is nonnegative, then the given equation requires that xβlogy=x+logy to that βlogy=logy=0 and y=1. On the other hand, if (xβlogy) is negative β(xβlogy)=x+logy so that 2x=0. We can write x(yβ1)=0 to any that x=0 or y=1 or both.