Problem:
What is the value of
(k=1β20βlog5kβ3k2)β
(k=1β100βlog9kβ25k)?
Answer Choices:
A. 21
B. 100log5β3
C. 200log3β5
D. 2,200
E. 21,000
Solution:
Note that
log5kβ3k2=log5klog3k2β=klog5k2log3β=log5log3ββ
k
and
log9kβ25k=log9klog25kβ=2klog32klog5β=log3log5β.
Thus
k=1β20βlog5kβ3k2=log5log3βk=1β20βk=log5log3ββ
220β
21β=210β
log5log3β
and
k=1β100βlog9kβ25k=100β
log3log5β
The requested product is 210β
100=21,000β.
The problems on this page are the property of the MAA's American Mathematics Competitions