Problem:
What is the value of
log37⋅log59⋅log711⋅log913⋯log2125⋅log2327?
Answer Choices:
A. 3
B. 3log723
C. 6
D. 9
E. 10
Solution:
The change of base formula states that logab=logalogb. Thus the product telescopes:
log3log7⋅log5log9⋅log7log11⋅log9log13⋯log21log25⋅log23log27=log3log25⋅log5log27=log3log52⋅log5log33=log32log5⋅log53log3=6.
OR
Let
a=log37⋅log711⋅log1115⋅log1519⋅log1923⋅log2327
and
b=log59⋅log913⋅log1317⋅log1721⋅log2125.
The required product is ab. Now
b=log59⋅log913⋅log1317⋅log1721⋅log2125=log59log913⋅log1317⋅log1721⋅log2125=log513⋅log1317⋅log1721⋅log2125=log513log1317⋅log1721⋅log2125=log517⋅log1721⋅log2125=log517log1721⋅log2125=log521⋅log2125=log521log2125=log525=2
Similarly, a=log327=3, so ab=2⋅3=6.
The problems on this page are the property of the MAA's American Mathematics Competitions