Problem:
The domain of the function f(x)=log21​​(log4​(log41​​(log16​(log161​​x)))) is an interval of length nm​, 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,aî€ =1, the domain of loga​x is the set {x:x>0}. Moreover, for 0<a<1,loga​x is a decreasing function on its domain, and for a>1,loga​x is an increasing function on its domain. Thus the function f(x) is defined if and only if log4​(log41​​(log16​(log161​​(x))))>0, and this inequality is
equivalent to each of the following:
log41​​(log16​(log161​​(x)))>1,0<log16​(log161​​x)<41​1<log161​​x<2, and 2561​<x<161​​
Thus nm​=161​−2561​=25615​, and m+n=271​.
The problems on this page are the property of the MAA's American Mathematics Competitions