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
logaba+logbab=logaa−logab+logbb−logba=1−logab+1−logba=2−logab−logba
Let c=logab, and note that c>0 since a and b are both greater than 1 . Thus
logaba+logbab=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
logaba+logbab=2−c−c1
From the Arithmetic-Geometric Mean Inequality we have
2c+1/c≥c⋅c1=1, so c+c1≥2
and
logaba+logbab=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