# Difference between revisions of "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 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}, 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 $x=2$, $x^{2^{\sqrt{2}}}>\sqrt{2}^{2^{x}}$, thus the answer cannot be $C$. 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 $\boxed{\textbf{(D) } 2 \le x < 6}$.