Problem:
What is the value of
log402log280−log202log2160?
Answer Choices:
A. 0
B. 1
C. 45
D. 2
E. log25
Solution:
Note that log402=log2401, and similarly log202=log2201
log402log280−log202log2160=log280⋅log240−log2160⋅log220=(log24+log220)(log22+log220)−(log28+log220)log220=(2+log220)(1+log220)−(3+log220)log220
Expanding,
2+2log220+log220+(log220)2−3log220−(log220)2
All the log terms cancel, so the answer is (D) 2 .
The problems on this page are the property of the MAA's American Mathematics Competitions