2021 AMC 12B Problems/Problem 21

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?

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

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

Video Solution by OmegaLearn (Logarithmic Tricks)

https://youtu.be/uCTpLB-kGR4

~ pi_is_3.14

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2021 AMC 12B Problems/Problem 21

Contents

Problem

Solution (Rough Approximation)

Video Solution by OmegaLearn (Logarithmic Tricks)

See Also