Difference between revisions of "2003 AMC 12A Problems/Problem 24"
m (→Solution) |
|||
(5 intermediate revisions by 2 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == 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> | If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math> | ||
Line 12: | Line 11: | ||
== Solution == | == Solution == | ||
− | |||
Using logarithmic rules, we see that | Using logarithmic rules, we see that | ||
Line 18: | Line 16: | ||
<cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b})</cmath> | <cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b})</cmath> | ||
− | Since <math>a</math> and <math>b</math> are both | + | Since <math>a</math> and <math>b</math> are both greater than <math>1</math>, 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=0 \Rightarrow \boxed{\textbf{B}}.</math> |
Note that the maximum occurs when <math>a=b</math>. | Note that the maximum occurs when <math>a=b</math>. | ||
Line 25: | Line 23: | ||
The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s | The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s | ||
+ | |||
+ | -MistyMathMusic | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2003|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2003|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:03, 19 April 2021
Contents
Problem
If what is the largest possible value of
Solution
Using logarithmic rules, we see that
Since and are both greater than , using AM-GM gives that the term in parentheses must be at least , so the largest possible values is
Note that the maximum occurs when .
Video Solution
The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s
-MistyMathMusic
See Also
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.