Problem:
For how many positive integers x is log10​(x−40)+log10​(60−x)<2 ?
Answer Choices:
A. 10
B. 18
C. 19
D. 20
E. infinitelymany
Solution:
The domain of log10​(x−40)+log10​(60−x) is 40<x<60. Within this domain, the inequality log10​(x−40)+log10​(60−x)<2 is equivalent to each of the following: log10​((x−40)(60−x))<2,(x−40)(60−x)< 102=100,x2−100x+2500>0, and (x−50)2>0. The last inequality is true for all xî€ =50. Thus the integer solutions to the original inequality are 41,42,…,49,51,52,…,59, and their number is 18​ .
The problems on this page are the property of the MAA's American Mathematics Competitions