Difference between revisions of "2021 AMC 12B Problems/Problem 21"
(→Video Solution by OmegaLearn (Logarithmic Tricks)) |
(→Solution (Rough Approximation)) |
||
| Line 12: | Line 12: | ||
Note that this solution is not recommended. | Note that this solution is not recommended. | ||
| − | 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>A</math> or <math>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>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{D}</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>A</math> or <math>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>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{D}</math>. |
==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 00:16, 12 February 2021
Contents
Problem
Let 
be the sum of all positive real numbers 
for which![\[x^{2^{\sqrt2}}=\sqrt2^{2^x}.\]](http://latex.artofproblemsolving.com/9/1/d/91d43f026978eb57bfbe0d83263ca432924008bf.png)
Which 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.
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 $x^(2^\sqrt{2})$ (Error compiling LaTeX. Unknown error_msg) (for greater values of ) and is greater than $x^(2^\sqrt{2})$ (Error compiling LaTeX. Unknown error_msg) for 
and less than $x^(2^\sqrt{2})$ (Error compiling LaTeX. Unknown error_msg) for 
, where 
is the second solution. Thus, the answer cannot be 
or 
. We then start plugging in numbers to roughly approximate the answer. When 
, $x^(2^\sqrt{2})>\sqrt{2}^(2^x)$ (Error compiling LaTeX. Unknown error_msg), thus the answer cannot be 
. Then, when 
, $x^(2^\sqrt{2})=4^(2^\sqrt{2})<64<\sqrt{2}^(2^x)=256$ (Error compiling LaTeX. Unknown error_msg). Therefore, 
, so the answer is 
.
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. 
