Difference between revisions of "2021 AMC 12B Problems/Problem 21"
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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>. | + | 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 |
==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 01:28, 12 February 2021
Contents
Problem
Let be the sum of all positive real numbers for whichWhich of the following statements is true?
Video Solution by OmegaLearn (Logarithmic Tricks)
~ pi_is_3.14
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
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 |
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