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

(my solution was incorrect)
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<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 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=0 \Rightarrow \boxed{\textbf{B}}.</math>
+
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>.

Latest revision as of 12:03, 19 April 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 greater than $1$, 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 (ProblemsAnswer KeyResources)
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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