Difference between revisions of "2021 AMC 12B Problems/Problem 21"
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<math>\textbf{(A) }S<\sqrt2 \qquad \textbf{(B) }S=\sqrt2 \qquad \textbf{(C) }\sqrt2<S<2\qquad \textbf{(D) }2\le S<6 \qquad \textbf{(E) }S\ge 6</math> | <math>\textbf{(A) }S<\sqrt2 \qquad \textbf{(B) }S=\sqrt2 \qquad \textbf{(C) }\sqrt2<S<2\qquad \textbf{(D) }2\le S<6 \qquad \textbf{(E) }S\ge 6</math> | ||
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+ | == Solution (Rough Approximation) == | ||
+ | Note that this solution is not recommended unless you're running out of time. | ||
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+ | Upon pure observation, it is obvious that one solution to this equality is <math>x=\sqrt{2}</math>. From this, we can deduce that this equality has two solutions, since <math>\sqrt{2}^{2^{x}}</math> grows faster than <math>x^{2^{\sqrt{2}}}</math> (for greater values of <math>x</math>) and <math>\sqrt{2}^{2^{x}}</math> is greater than <math>x^{2^{\sqrt{2}}}</math> for <math>x<\sqrt{2}</math> and less than <math>x^{2^{\sqrt{2}}}</math> for <math>\sqrt{2}<x<n</math>, where <math>n</math> is the second solution. Thus, the answer cannot be <math>\text{A}</math> or <math>\text{B}</math>. We then start plugging in numbers to roughly approximate the answer. When <math>x=2</math>, <math>x^{2^{\sqrt{2}}}>\sqrt{2}^{2^{x}}</math>, thus the answer cannot be <math>\text{C}</math>. Then, when <math>x=4</math>, <math>x^{2^{\sqrt{2}}}=4^{2^{\sqrt{2}}}<64<\sqrt{2}^{2^{x}}=256</math>. Therefore, <math>S<4+\sqrt{2}<6</math>, so the answer is <math>\boxed{\textbf{(D) } 2 \le S < 6}</math>. ~Baolan | ||
== Video Solution by OmegaLearn (Logarithmic Tricks) == | == Video Solution by OmegaLearn (Logarithmic Tricks) == | ||
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~ pi_is_3.14 | ~ pi_is_3.14 | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=20|num-a=22}} | {{AMC12 box|year=2021|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:14, 12 February 2021
Contents
[hide]Problem
Let be the sum of all positive real numbers for whichWhich of the following statements is true?
Solution (Rough Approximation)
Note that this solution is not recommended unless you're running out of time.
Upon pure observation, it is obvious that one solution to this equality is . From this, we can deduce that this equality has two solutions, since grows faster than (for greater values of ) and is greater than for and less than for , where is the second solution. Thus, the answer cannot be or . We then start plugging in numbers to roughly approximate the answer. When , , thus the answer cannot be . Then, when , . Therefore, , so the answer is . ~Baolan
Video Solution by OmegaLearn (Logarithmic Tricks)
~ pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.