2003 AMC 12A Problems/Problem 24

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

Solution 1

Using logarithmic rules, we see that

$\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)$ $=2-(\log_{a}b+\frac {1}{\log_{a}b})$

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 $a=b$ .

Solution 2

We arrive at the expression $2 - \left(\log_a b + \log_b a\right)$ the same way as the previous solution. However, there is a logarithm property that states that $\log_a b = -\log_b a$ (this should come intuitively, you can try it with an example). This means that $\log_a b + \log_b a = 0$ and thus the expression in the problem statement simplifies conveniently to $2$ . This is the largest (and smallest) value possible, so $2 \Rightarrow \boxed{\textbf{B}}$ is the answer.

Video Solution

The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s

-MistyMathMusic

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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