Difference between revisions of "2022 AMC 10A Problems/Problem 11"
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We can rewrite the equation using fractional exponents and take logarithms of both sides: <cmath>\log_2{(2^{m}\cdot4096^{-1/2}}) = \log_2{(2\cdot4096^{-1/m})}.</cmath> | We can rewrite the equation using fractional exponents and take logarithms of both sides: <cmath>\log_2{(2^{m}\cdot4096^{-1/2}}) = \log_2{(2\cdot4096^{-1/m})}.</cmath> | ||
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+ | ==Solution 3== | ||
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+ | Since surd roots are conventionally positive integers, assume <math>m</math> is an integer, so <math>m</math> can only be <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>6</math>, and <math>12</math>. <math>\sqrt{\frac{1}{4096}}=\frac{1}{64}</math>. Testing out <math>m</math>, we see that only <math>3</math> and <math>4</math> work. Hence, <math>3+4=\boxed{\textbf{(C) }7}</math>. | ||
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+ | ~MrThinker | ||
==Video Solution 1 == | ==Video Solution 1 == |
Revision as of 14:15, 14 October 2023
Contents
[hide]Problem
Ted mistakenly wrote as What is the sum of all real numbers for which these two expressions have the same value?
Solution 1
We are given that Converting everything into powers of we have We multiply both sides by , then rearrange as By Vieta's Formulas, the sum of such values of is
Note that or from the quadratic equation above.
~MRENTHUSIASM
~KingRavi
Solution 2 (Logarithms)
We can rewrite the equation using fractional exponents and take logarithms of both sides: We can then use the additive properties of logarithms to split them up: Using the power rule, the fact that and bringing the exponents down, we get from which or . Therefore, the answer is
- abed_nadir
Solution 3
Since surd roots are conventionally positive integers, assume is an integer, so can only be , , , , , and . . Testing out , we see that only and work. Hence, .
~MrThinker
Video Solution 1
~Education, the Study of Everything
Video Solution (Easy)
~Whiz
See Also
2022 AMC 10A (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.