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Difference between revisions of "2003 AMC 12A Problems/Problem 24"

(Replaced content with ' == See Also == {{AMC12 box|year=2003|ab=A|num-b=23|after=25}}')
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== Problem ==
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If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math>
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<math>
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\mathrm{(A)}\ -2      \qquad
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\mathrm{(B)}\ 0    \qquad
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\mathrm{(C)}\ 2      \qquad
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\mathrm{(D)}\ 3      \qquad
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\mathrm{(E)}\ 4
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</math>
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== Solution ==
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Using logarithmic rules, we see that
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<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath>
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<cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b})</cmath>
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Since <math>a</math> and <math>b</math> are both positive, using [[AM-GM]] gives that the term in parentheses must be at least <math>2</math>, so the largest possible values is <math>2-2=\boxed{0}.</math>
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2003|ab=A|num-b=23|after=25}}
 
{{AMC12 box|year=2003|ab=A|num-b=23|after=25}}

Revision as of 13:55, 28 February 2010

Problem

If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$

$\mathrm{(A)}\ -2 \qquad \mathrm{(B)}\ 0 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 3 \qquad \mathrm{(E)}\ 4$

Solution

Using logarithmic rules, we see that

\[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\] \[=2-(\log_{a}b+\frac {1}{\log_{a}b})\]

Since $a$ and $b$ are both positive, using AM-GM gives that the term in parentheses must be at least $2$, so the largest possible values is $2-2=\boxed{0}.$

See Also

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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