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

Revision as of 12:55, 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 $2-2=\boxed{0}.$

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

@@ Line 1: / Line 1: @@
+== Problem ==
+If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math>
+<math>
+\mathrm{(A)}\ -2      \qquad
+\mathrm{(B)}\ 0     \qquad
+\mathrm{(C)}\ 2      \qquad
+\mathrm{(D)}\ 3      \qquad
+\mathrm{(E)}\ 4
+</math>
+== Solution ==
+Using logarithmic rules, we see that
+<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath>
+<cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b})</cmath>
+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 ==
 {{AMC12 box|year=2003|ab=A|num-b=23|after=25}}

Page

Toolbox

Search

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

Revision as of 12:55, 28 February 2010

Problem

Solution

See Also