Problem:
What is the value of a for which log2a1+log3a1+log4a1=1 ?
Answer Choices:
A. 9
B. 12
C. 18
D. 24
E. 36
Solution:
By the change of base formula, logmn1=lognm. Thus
1=log2a1+log3a1+log4a1=loga2+loga3+loga4=loga24
It follows that a=24.
The problems on this page are the property of the MAA's American Mathematics Competitions