2021 AMC 12B Problems/Problem 21

Revision as of 01:14, 12 February 2021 by Baolan (talk | contribs) (Video Solution by OmegaLearn (Logarithmic Tricks))

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 $\sqrt{2}^2^x$ (Error compiling LaTeX. ! Double superscript.) grows faster than $x^2^\sqrt{2}$ (Error compiling LaTeX. ! Double superscript.) (for greater values of $x$) and $\sqrt{2}^2^x$ (Error compiling LaTeX. ! Double superscript.) is greater than $x^2^\sqrt{2}$ (Error compiling LaTeX. ! Double superscript.) for $x<\sqrt{2}$ and less than $x^2^\sqrt{2}$ (Error compiling LaTeX. ! Double superscript.) 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. ! Double superscript.), 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. ! Double superscript.). Therefore, $S<4+\sqrt{2}<6$, so the answer is $\boxed{D}$.

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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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