Problem: For all positive numbers x distinct from 1,
log3x1+log4x1+log5x1
equals
Answer Choices:
A. log60x1
B. logx601
C. (log3x)(log4x)(log5x)1
D. (log3x)+(log4x)+(log5x)12
E. (log3x)(log5x)log2x+(log2x)(log5x)log3x+(log2x)(log3x)log5x
Solution:
If y=logab, then ay=b and a=b1/y. Thus
logba=y1=logab1
Thereiore,
log3x1+log4x1+log5x1=logx3+logx4+logx5=logx60=log60xi.