Problem:
For how many positive integers x is log10β(xβ40)+log10β(60βx)<2 ?
Answer Choices:
A. 10
B. 18
C. 19
D. 20
E. infinitelymany
Solution:
The domain of log10β(xβ40)+log10β(60βx) is 40<x<60. Within this domain, the inequality log10β(xβ40)+log10β(60βx)<2 is equivalent to each of the following: log10β((xβ40)(60βx))<2,(xβ40)(60βx)< 102=100,x2β100x+2500>0, and (xβ50)2>0. The last inequality is true for all xξ =50. Thus the integer solutions to the original inequality are 41,42,β¦,49,51,52,β¦,59, and their number is 18β .
The problems on this page are the property of the MAA's American Mathematics Competitions