Difference between revisions of "2022 AMC 12A Problems/Problem 11"
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First, notice that there must be two such numbers: one greater than <math>\log_69</math> and one less than it. Furthermore, they both have to be the same distance away, namely <math>2(\log_610 - 1)</math>. Let these two numbers be <math>\log_6a</math> and <math>\log_6b</math>. Because they are equidistant from <math>\log_69</math>, we have <math>\frac{\log_6a + \log_6b}{2} = \log_69</math>. Using log properties, this simplifies to <math>\log_6{\sqrt{ab}} = \log_69</math>. We then have <math>\sqrt{ab} = 9</math>, so <math>ab = \boxed{\textbf{(E) } 81}</math>. | First, notice that there must be two such numbers: one greater than <math>\log_69</math> and one less than it. Furthermore, they both have to be the same distance away, namely <math>2(\log_610 - 1)</math>. Let these two numbers be <math>\log_6a</math> and <math>\log_6b</math>. Because they are equidistant from <math>\log_69</math>, we have <math>\frac{\log_6a + \log_6b}{2} = \log_69</math>. Using log properties, this simplifies to <math>\log_6{\sqrt{ab}} = \log_69</math>. We then have <math>\sqrt{ab} = 9</math>, so <math>ab = \boxed{\textbf{(E) } 81}</math>. | ||
Revision as of 16:45, 15 October 2023
Contents
[hide]Problem
What is the product of all real numbers such that the distance on the number line between and is twice the distance on the number line between and ?
Solution 1
Let .
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~ oinava
Solution 2
First, notice that there must be two such numbers: one greater than and one less than it. Furthermore, they both have to be the same distance away, namely . Let these two numbers be and . Because they are equidistant from , we have . Using log properties, this simplifies to . We then have , so .
~ jamesl123456
Solution 2 (Logarithmic Rules and Casework)
In effect we must find all such that where .
Notice that by log rules Using log rules again,
Now we proceed by casework for the distinct values of .
Case 1
Subbing in for and using log rules, From this we may conclude that
Case 2
Subbing in for and using log rules, From this we conclude that
Finding the product of the distinct values,
~Spektrum
Video Solution 1 (Quick and Simple)
~Education, the Study of Everything
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.