2003 AMC 12A Problems/Problem 24: Difference between revisions
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== 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> | ||
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== Solution == | == Solution == | ||
=== Solution 1 === | |||
Using logarithmic rules, we see that | Using logarithmic rules, we see that | ||
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Note that the maximum occurs when <math>a=b</math>. | Note that the maximum occurs when <math>a=b</math>. | ||
=== Solution 2 === | |||
We arrive at the expression <math>2 - \left(\log_a b + \log_b a\right)</math> the same way as the previous solution. However, there is a logarithm property that states that <math>\log_a b = -\log_b a</math> (this should come intuitively, you can try it with an example). This means that <math>\log_a b + \log_b a = 0</math> and thus the expression in the problem statement simplifies conveniently to <math>2</math>. This is the largest (and smallest) value possible, so <math>2 \Rightarrow \boxed{\textbf{B}}</math> is the answer. | |||
==Video Solution== | ==Video Solution== | ||
Revision as of 01:33, 25 January 2021
Problem
If 
what is the largest possible value of 
Solution
Solution 1
Using logarithmic rules, we see that
![\[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\]](http://latex.artofproblemsolving.com/a/c/8/ac81a63197833883d0f23b2c94e96af4303e9c3e.png)
![\[=2-(\log_{a}b+\frac {1}{\log_{a}b})\]](http://latex.artofproblemsolving.com/8/6/9/86936dda057f6fa26fa792fcec247eb7fd350936.png)
Since 
and 
are both positive, 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 
.
Solution 2
We arrive at the expression 
the same way as the previous solution. However, there is a logarithm property that states that 
(this should come intuitively, you can try it with an example). This means that 
and thus the expression in the problem statement simplifies conveniently to . This is the largest (and smallest) value possible, so 
is the answer.
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 | |
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