# 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

## Solution 1 (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 $\text{A}$ 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

## Solution 2

Note that \begin{align*} x^{2^{\sqrt{2}}} &= {\sqrt{2}}^{2^x} \\ 2^{\sqrt{2}} \log_2 x &= 2^{x} \log_2 \sqrt{2}. \end{align*} (At this point we see by inspection that $x=\sqrt{2}$ is a solution.)

We simplify the RHS, then take the base- $2$ logarithm for both sides: \begin{align*} 2^{\sqrt{2}} \log_2 x &= 2^{x-1} \\ \log_2{\left(2^{\sqrt{2}} \log_2 x\right)} &= x-1 \\ \sqrt{2} + \log_2 \log_2 x &= x-1 \\ \log_2 \log_2 x &= x - 1 - \sqrt{2}. \end{align*} The RHS is a line; the LHS is a concave curve that looks like a logarithm and has $x$ intercept at $(2,0).$

There are at most two solutions, one of which is $\sqrt{2}.$ But note that at $x=2,$ we have $\log_2 \log_2 {2} = 0 > 2 - 1 - \sqrt{2},$ meaning that the log log curve is above the line, so it must intersect the line again at a point $x > 2.$ Now we check $x=4$ and see that $\log_2 \log_2 {4} = 1 < 4 - 1 - \sqrt{2},$ which means at $x=4$ the line is already above the log log curve. Thus, the second solution lies in the interval $(2,4).$ The answer is $\boxed{\textbf{(D) }2\le S<6}.$

~ ccx09

## Solution 3 (Graphs and Light Approximations)

We rewrite the right side, then take the base- $2$ logarithm for both sides: \begin{align*} x^{2^{\sqrt2}}&=\left(2^\frac12\right)^{2^x} \\ x^{2^{\sqrt2}}&=2^{\frac12\cdot2^x} \\ x^{2^{\sqrt2}}&=2^{2^{x-1}} \\ \log_2{\left(x^{2^{\sqrt2}}\right)}&=\log_2{\left(2^{2^{x-1}}\right)} \\ 2^{\sqrt2}\log_2{x}&=2^{x-1}. \hspace{20mm} (*) \end{align*} By observations, $x=\sqrt2$ is one solution. By quick sketches of $f(x)=2^{\sqrt2}\log_2{x}$ and $g(x)=2^{x-1},$ we know that $(*)$ has two solutions, with $x=\sqrt2$ the smaller solution. We construct the following table of values: $$\begin{array}{c|c|c|c} & & & \\ [-2ex] \boldsymbol{x} & \boldsymbol{f(x)=2^{\sqrt2}\log_2{x}} & \boldsymbol{g(x)=2^{x-1}} & \textbf{Comparison} \\ [1ex] \hline & & & \\ [-1ex] 1 & 0 & 1 & \\ [1ex] \sqrt2 & 2^{\sqrt2-1} & 2^{\sqrt2-1} & f\left(\sqrt2\right)=g\left(\sqrt2\right) \\ [1ex] 2 & 2^{\sqrt2} & 2 & f(2)>g(2) \\ [1ex] 3 & 2^{\sqrt2}\log_2{3} & 2^2 & \\ [1ex] 4 & 2^{\sqrt2+1} & 2^3 & f(4) Let $x=t$ be the larger solution. Since exponential functions outgrow logarithmic functions, we have $f(x) for all $x>t.$ By the Intermediate Value Theorem, we get $t\in(2,4),$ from which $\[S=\sqrt2+t\in\left(\sqrt2+2,\sqrt2+4\right).$$ Finally, approximating with $\sqrt2\approx1.414$ results in $\boxed{\textbf{(D) }2\le S<6}.$

Graphs of $f(x)$ and $g(x)$ in Desmos: https://www.desmos.com/calculator/xyxzsvjort

~MRENTHUSIASM

~ pi_is_3.14

## Video Solution by hippopotamus1

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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