Difference between revisions of "2003 AMC 12A Problems/Problem 24"
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== Problem == | == Problem == | ||
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If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math> | If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math> | ||
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== Solution == | == Solution == | ||
+ | Using logarithmic rules, we see that | ||
+ | |||
+ | <cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath> | ||
+ | <cmath>=2-\left(\log_{a}b+\frac {1}{\log_{a}b}\right)</cmath> | ||
+ | |||
+ | Since <math>a</math> and <math>b</math> are both greater than <math>1</math>, using [[AM-GM]] gives that the term in parentheses must be at least <math>2</math>, so the largest possible values is <math>2-2=0 \Rightarrow \boxed{\textbf{B}}.</math> | ||
+ | |||
+ | 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> | <cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath> | ||
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− | + | 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 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right)</cmath>. | |
+ | |||
+ | We can now cross multiply to get that <cmath>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)</cmath> Finally, we cancel to get <math>2-2=0 \Rightarrow \boxed{\textbf{(B) 0}}.</math> | ||
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+ | Solution by: armang32324 | ||
+ | |||
+ | ==Solution 3 (Calculus)== | ||
+ | By the logarithmic rules, we have <math>2-(\log_a{b}+\frac{1}{\log_a{b}})</math>. | ||
+ | Let <math>x=\log_a{b}</math>. Thus, the expression becomes <math>2-(x+\frac{1}{x})</math>. 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: <math>\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.</math> | ||
+ | Since <math>a\geq b>1, x=-1</math> is not a solution. Thus, <math>x=1</math>. Substituting it into the original expression <math>2-(x+\frac{1}{x})</math>, we get <math>2-(1+\frac{1}{1})=2-2=\boxed{0}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s | ||
+ | |||
+ | -MistyMathMusic | ||
== See Also == | == See Also == | ||
− | + | {{AMC12 box|year=2003|ab=A|num-b=23|num-a=25}} | |
− | {{AMC12 box|year=2003|ab=A|num-b=23| | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 08:35, 6 September 2023
Contents
Problem
If what is the largest possible value of
Solution
Using logarithmic rules, we see that
Since and are both greater than , using AM-GM gives that the term in parentheses must be at least , so the largest possible values is
Note that the maximum occurs when .
Solution 2 (More Algebraic)
Similar to the previous solution, we use our logarithmic rules and start off the same way.
However, now, we use a different logarithmic rule stating that is simply equal to . With this, we can rewrite our previous equation to give us .
We can now cross multiply to get that Finally, we cancel to get
Solution by: armang32324
Solution 3 (Calculus)
By the logarithmic rules, we have . Let . Thus, the expression becomes . 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: Since is not a solution. Thus, . Substituting it into the original expression , we get .
Video Solution
The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s
-MistyMathMusic
See Also
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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