Difference between revisions of "2022 AMC 12A Problems/Problem 11"
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<math>\textbf{(A) } 10 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 25 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 81</math> | <math>\textbf{(A) } 10 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 25 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 81</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | 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)} | + | 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> 9b^1 \cdot 9b^{-1} = \boxed{81}</math>. | ||
+ | |||
+ | ~ oinava | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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>. | ||
~ jamesl123456 | ~ jamesl123456 | ||
− | ==Solution | + | ==Solution 3 (Logarithmic Rules and 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>. | 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>. | ||
Line 16: | Line 25: | ||
Notice that by log rules | Notice that by log rules | ||
<cmath> | <cmath> | ||
− | d = \log_6 10 - 1 = log_6 \frac{10}{6} | + | d = \log_6 10 - 1 = \log_6 \frac{10}{6} |
</cmath> | </cmath> | ||
Using log rules again, | Using log rules again, | ||
Line 52: | Line 61: | ||
Finding the product of the distinct values, | Finding the product of the distinct values, | ||
− | <math>x_1x_2 = \boxed{\textbf{(E)} | + | <math>x_1x_2 = \boxed{\textbf{(E) } 81}</math> |
~Spektrum | ~Spektrum | ||
Line 61: | Line 70: | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
+ | ==Video Solution 1 (Understand the question first)== | ||
+ | https://youtu.be/7yAh4MtJ8a8?si=CsVGUSiyUiT4nNI0&t=2076 | ||
+ | |||
+ | ~Math-X | ||
== 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}} |
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 |
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