Problem:
The domain of the function f(x)=log21ββ(log4β(log41ββ(log16β(log161ββx)))) is an interval of length nmβ, where m and n are relatively prime positive integers. What is m+n?
Answer Choices:
A. 19
B. 31
C. 271
D. 319
E. 511
Solution:
For every a>0,aξ =1, the domain of logaβx is the set {x:x>0}. Moreover, for 0<a<1,logaβx is a decreasing function on its domain, and for a>1,logaβx is an increasing function on its domain. Thus the function f(x) is defined if and only if log4β(log41ββ(log16β(log161ββ(x))))>0, and this inequality is
equivalent to each of the following:
log41ββ(log16β(log161ββ(x)))>1,0<log16β(log161ββx)<41β1<log161ββx<2, and 2561β<x<161ββ
Thus nmβ=161ββ2561β=25615β, and m+n=271β.
The problems on this page are the property of the MAA's American Mathematics Competitions