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 1

Using logarithmic rules, we see that

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

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 $a=b$ .

Solution 2 (Calculus)

By the logarithmic rules, we have $2-(\log_a{b}+\frac{1}{\log_a{b}})$ . Let $x=\log_a{b}$ . Thus, the expression becomes $2-(x+\frac{1}{x})$ . We want to find the maximum of the function. To do so, we will find its derivative and let it equal to 0. We get: $\frac{d}{dx}\big(2-(x+\frac{1}{x})\big)=\frac{d}{dx}\big(2-x-\frac{1}{x})=-1+x^{-2}=0 \implies \frac{1}{x^2}=1, x^2=1, x=\pm1.$ Since $a\geq b>1, x=-1$ is not a solution. Thus, $x=1$ . Substituting it into the original expression $2-(x+\frac{1}{x})$ , we get $2-(1+\frac{1}{1})=2-2=\boxed{0}$ .

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

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2003 AMC 12A Problems/Problem 24

Contents

Problem

Solution 1

Solution 2 (Calculus)

Video Solution

See Also