Difference between revisions of "2022 AMC 12A Problems/Problem 11"
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==Solution 1== | ==Solution 1== | ||
− | Let <math>a = 2 \cdot |\log_6 10 - 1| = |\log_6 9 - \log_6 x| = |\log_6 \frac{9}{x}| </math>. | + | Let <math>a = 2 \cdot |\log_6 10 - 1| = |\log_6 9 - \log_6 x| = \left|\log_6 \frac{9}{x}\right| </math>. |
<math> \pm a = \log_6 \frac{9}{x} \implies 6^{\pm a} = b^{\pm 1} = \frac{9}{x} \implies x = 9 \cdot b^{\pm 1} </math> | <math> \pm a = \log_6 \frac{9}{x} \implies 6^{\pm a} = b^{\pm 1} = \frac{9}{x} \implies x = 9 \cdot b^{\pm 1} </math> |
Latest revision as of 01:00, 25 September 2024
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 .
.
~ 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 3 (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
Video Solution 1 (Understand the question first)
https://youtu.be/7yAh4MtJ8a8?si=CsVGUSiyUiT4nNI0&t=2076
~Math-X
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.