Problem:
If a≥b>1, what is the largest possible value of loga​(a/b)+logb​(b/a) ?
Answer Choices:
A. −2
B. 0
C. 2
D. 3
E. 4
Solution:
We have
loga​ba​+logb​ab​=loga​a−loga​b+logb​b−logb​a=1−loga​b+1−logb​a=2−loga​b−logb​a​
Let c=loga​b, and note that c>0 since a and b are both greater than 1 . Thus
loga​ba​+logb​ab​=2−c−c1​=−cc2−2c+1​=−c(c−1)2​≤0
This expression is 0​ when c=1, that is, when a=b.
\section*{OR}
As above
loga​ba​+logb​ab​=2−c−c1​
From the Arithmetic-Geometric Mean Inequality we have
2c+1/c​≥c⋅c1​​=1, so c+c1​≥2
and
loga​ba​+logb​ab​=2−(c+c1​)≤0
with equality when c=c1​, that is, when a=b.
The problems on this page are the property of the MAA's American Mathematics Competitions