# 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

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

-MistyMathMusic

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