Difference between revisions of "2022 AMC 12B Problems/Problem 16"
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Substitution into <math>(\log_2{x})^{\log_2{y}}=2^{7}</math> yields | Substitution into <math>(\log_2{x})^{\log_2{y}}=2^{7}</math> yields | ||
− | <math>(\dfrac{64}{y})^{\log_2{y}}=2^{7} \Rightarrow \log_2{y}\log_2{\dfrac{64}{y}}=7 \Rightarrow \log_2{y}(6-\log_2{y})=7 \Rightarrow \log^2_2{y}-6\log_2{y}+7=0</math>. | + | <math>\left(\dfrac{64}{y}\right)^{\log_2{y}}=2^{7} \Rightarrow \log_2{y}\log_2{\dfrac{64}{y}}=7 \Rightarrow \log_2{y}(6-\log_2{y})=7 \Rightarrow \log^2_2{y}-6\log_2{y}+7=0</math>. |
Solving for <math>\log_2{y}</math> yields <math>\log_2{y}=3-\sqrt{2}</math> or <math>3+\sqrt{2}</math>, and we take the greater value <math>\boxed{\boldsymbol{(\textbf{C})3+\sqrt{2}}}</math>. | Solving for <math>\log_2{y}</math> yields <math>\log_2{y}=3-\sqrt{2}</math> or <math>3+\sqrt{2}</math>, and we take the greater value <math>\boxed{\boldsymbol{(\textbf{C})3+\sqrt{2}}}</math>. |
Latest revision as of 23:10, 1 November 2024
Contents
[hide]Problem
Suppose and are positive real numbers such that What is the greatest possible value of ?
Solution
Take the base-two logarithm of both equations to get Now taking the base-two logarithm of the first equation again yields It follows that the real numbers and satisfy and . Solving this system yields Thus the largest possible value of is .
cr. djmathman
Solution 2
.
Substitution into yields
.
Solving for yields or , and we take the greater value .
~4SunnyH
Solution 3
Let We have and .
Then, from eq 1, and substituting in to eq 2, Thus,
Solving for using the quadratic formula gets Since we are looking for which equals we put as our answer.
~sirswagger21
Video Solution by mop 2024
https://youtu.be/ezGvZgBLe8k&t=722s
~r00tsOfUnity
Video Solution (Just 2 min!)
Education, the Study of Everything
Video Solution(1-16)
~~Hayabusa1
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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.