2003 AMC 12A Problems/Problem 24: Difference between revisions

Revision as of 19:35, 6 April 2023

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 $a=b$ .

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 $2-(\log_{a}b+\log_{b}a) = 2 - (\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b})$ .

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

Solution by: armang32324

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

@@ Line 19: / Line 19: @@
 Note that the maximum occurs when <math>a=b</math>.
+==Solution 2 (More Algebraic)==
+Similar to the previous solution, we use our logarithmic rules and start off the same way.
+<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath>
+However, now, we use a different logarithmic rule stating that <math>\log_{a}b</math> is simply equal to <math>\frac {\log_{10}b}{\log_{10}a}</math>. With this, we can rewrite our previous equation to give us <cmath>2-(\log_{a}b+\log_{b}a) = 2 - (\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b})</cmath>.
+We can now cross multiply to get that <cmath>2 - (\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}) = 2 - (\frac {2 * log_{10}b * \log_{10}a}{log_{10}b * \log_{10}a})</cmath> Finally, we cancel to get <math>2-2=0 \Rightarrow \boxed{\textbf{(B) 0}}.</math>
+Solution by: armang32324
 ==Video Solution==

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