Difference between revisions of "2022 AMC 12A Problems/Problem 11"
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~ jamesl123456 | ~ jamesl123456 | ||
+ | |||
+ | ==Solution 2(Log Rules + Casework)== | ||
+ | |||
+ | In effect we must find all <math>x</math> such that <math>\left|\log_6 9 - \log_6 x\right| = 2d</math> where <math>d = \log_6 10 - 1</math>. | ||
+ | |||
+ | Notice that by log rules | ||
+ | <cmath> | ||
+ | d = \log_6 10 - 1 = log_6 \frac{10}{6} | ||
+ | </cmath> | ||
+ | Using log rules again, | ||
+ | <cmath> | ||
+ | 2d = 2\log_6 \frac{10}{6} = \log_6 \frac{25}{9} | ||
+ | </cmath> | ||
+ | |||
+ | Now we proceed by casework for the distinct values of <math>x</math>. | ||
+ | |||
+ | ===Case 1=== | ||
+ | <cmath> | ||
+ | \log_6 9 - \log_6 x_1 = 2d | ||
+ | </cmath> | ||
+ | Subbing in for <math>2d</math> and using log rules, | ||
+ | <cmath> | ||
+ | \log_6 \frac{9}{x_1} = \log_6 \frac{25}{9} | ||
+ | </cmath> | ||
+ | From this we may conclude that | ||
+ | <cmath> | ||
+ | \frac{9}{x_1} = \frac{25}{9} \implies x_1 = \frac{81}{25} | ||
+ | </cmath> | ||
+ | |||
+ | ===Case 2=== | ||
+ | <cmath> | ||
+ | \log_6 9 - \log_6 x_2 = -2d | ||
+ | </cmath> | ||
+ | Subbing in for <math>-2d</math> and using log rules, | ||
+ | <cmath> | ||
+ | \log_6 \frac{9}{x_2} = \log_6 \frac{9}{25} | ||
+ | </cmath> | ||
+ | From this we conclude that | ||
+ | <cmath> | ||
+ | \frac{9}{x_2} = \frac{9}{25} \implies x_2 = 25 | ||
+ | </cmath> | ||
+ | |||
+ | Finding the product of the distinct values, | ||
+ | <math>x_1x_2 = \boxed{\textbf{(E)} \, 81}</math> | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2022|ab=A|num-b=10|num-a=12}} | {{AMC12 box|year=2022|ab=A|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:38, 13 November 2022
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
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(Log Rules + 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,
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.