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 1

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 2 (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: $\begin{array}{c|c|c|c} & & & \\ [-2ex] \boldsymbol{x} & \boldsymbol{f(x)=2^{\sqrt2}\log_2{x}} & \boldsymbol{g(x)=2^{x-1}} & \textbf{Comparison} \\ [1ex] \hline & & & \\ [-1ex] 1 & 0 & 1 & \\ [1ex] \sqrt2 & 2^{\sqrt2-1} & 2^{\sqrt2-1} & f\left(\sqrt2\right)=g\left(\sqrt2\right) \\ [1ex] 2 & 2^{\sqrt2} & 2 & f(2)>g(2) \\ [1ex] 3 & 2^{\sqrt2}\log_2{3} & 2^2 & \\ [1ex] 4 & 2^{\sqrt2+1} & 2^3 & f(4)<g(4) \\ [1ex] \end{array}$ Let $x=t$ be the larger solution. Since exponential functions outgrow logarithmic functions, we have $f(x)<g(x)$ 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

Solution 3 (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

Video Solution by hippopotamus1

https://www.youtube.com/watch?v=GjO6C_qC13U&feature=youtu.be

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