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 \cdot (\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> | + | 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 \cdot (\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>. |
~ jamesl123456 | ~ jamesl123456 |
Revision as of 12:55, 12 November 2022
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
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
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 |
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