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

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== Solution ==
 
== Solution ==
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=== 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>.
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=== Solution 2 ===
 
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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 01:35, 25 January 2021

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 $2-2=0 \Rightarrow \boxed{\textbf{B}}.$

Note that the maximum occurs when $a=b$.

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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