Problem: If f(x)=log(1βx1+xβ) for β1<x<1, then f(1+3x23x+x3β) in terms of f(x) is
Answer Choices:
A. βf(x)
B. 2f(x)
C. 3f(x)
D. [f(x)]2
E. [f(x)]3βf(x)
Solution:
f(1+3x23x+x3β)=log1β1+3x23x+x3β1+1+3x23x+x3ββ=log1+3x2β3xβx31+3x2+3x+x3β=log(1βx)3(1+x)3β=log(1βx1+xβ)3=3log1βx1+xβ=3f(x)