2003 AMC 12A Problems/Problem 24


If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$

$\mathrm{(A)}\ -2      \qquad \mathrm{(B)}\ 0     \qquad \mathrm{(C)}\ 2      \qquad \mathrm{(D)}\ 3      \qquad \mathrm{(E)}\ 4$


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 (More Algebraic)

Similar to the previous solution, we use our logarithmic rules and start off the same way. \[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\]

However, now, we use a different logarithmic rule stating that $\log_{a}b$ is simply equal to $\frac {\log_{10}b}{\log_{10}a}$. With this, we can rewrite our previous equation to give us \[2-(\log_{a}b+\log_{b}a) = 2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right)\].

We can now cross multiply to get that \[2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right) = 2 - \left(\frac {2 \cdot \log_{10}b \cdot \log_{10}a}{\log_{10}b \cdot \log_{10}a}\right)\] Finally, we cancel to get $2-2=0 \Rightarrow \boxed{\textbf{(B) 0}}.$

Solution by: armang32324

Solution 3 (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


See Also

2003 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AMC 12 Problems and Solutions

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