Difference between revisions of "2022 AMC 10A Problems/Problem 11"
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==Solution 3 (Logarithms)== | ==Solution 3 (Logarithms)== | ||
− | We can rewrite the equation using fractional exponents and take logarithms of both sides: | + | 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 then use the additive properties of logarithms to split them up: <cmath>\log_2{(2^{m})} + \log_2{(4096^{-1/2})} = \log_2{2} + \log_2{(4096^{-1/m})}.</cmath> | ||
+ | Using the power rule, the fact that <math>4096 = 2^{12},</math> and bringing the exponents down, we get | ||
+ | <cmath>\begin{align*} | ||
+ | m - 6 &= 1 - \frac{12}{m} \ | ||
+ | m + \frac{12}{m} &= 7 \ | ||
+ | m^{2} + 12 &= 7m \ | ||
+ | m^{2} - 7m + 12 &= 0 \ | ||
+ | (m-3)(m-4) &= 0, | ||
+ | \end{align*}</cmath> | ||
+ | from which <math>m = 3</math> or <math>m = 4</math>. | ||
− | + | Therefore, the answer is <math>3+4 = \boxed{\textbf{(C) } 7}.</math> | |
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- abed_nadir | - abed_nadir |
Revision as of 14:14, 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
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
Solution 3 (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
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.