# 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

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