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Difference between revisions of "2021 AMC 12B Problems/Problem 21"

(→‎Video Solution by OmegaLearn (Logarithmic Tricks))
(→‎Solution (Rough Approximation))
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Note that this solution is not recommended.
 
Note that this solution is not recommended.
  
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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>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>.
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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>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 01:16, 12 February 2021

Problem

Let $S$ be the sum of all positive real numbers $x$ for which\[x^{2^{\sqrt2}}=\sqrt2^{2^x}.\]Which of the following statements is true?

$\textbf{(A) }S<\sqrt2 \qquad \textbf{(B) }S=\sqrt2 \qquad \textbf{(C) }\sqrt2<S<2\qquad \textbf{(D) }2\le S<6 \qquad \textbf{(E) }S\ge 6$

Video Solution by OmegaLearn (Logarithmic Tricks)

https://youtu.be/uCTpLB-kGR4

~ 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 $x=\sqrt{2}$. From this, we can deduce that this equality has two solutions, since $\sqrt{2}^(2^x)$ grows faster than $x^(2^\sqrt{2})$ (Error compiling LaTeX. Unknown error_msg) (for greater values of $x$) and $\sqrt{2}^(2^x)$ is greater than $x^(2^\sqrt{2})$ (Error compiling LaTeX. Unknown error_msg) for $x<\sqrt{2}$ and less than $x^(2^\sqrt{2})$ (Error compiling LaTeX. Unknown error_msg) for $\sqrt{2}<x<n$, where $n$ is the second solution. Thus, the answer cannot be $A$ or $B$. We then start plugging in numbers to roughly approximate the answer. When $x=2$, $x^(2^\sqrt{2})>\sqrt{2}^(2^x)$ (Error compiling LaTeX. Unknown error_msg), thus the answer cannot be $C$. Then, when $x=4$, $x^(2^\sqrt{2})=4^(2^\sqrt{2})<64<\sqrt{2}^(2^x)=256$ (Error compiling LaTeX. Unknown error_msg). Therefore, $S<4+\sqrt{2}<6$, so the answer is $\boxed{D}$.

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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