Difference between revisions of "2021 AMC 12B Problems/Problem 21"

Revision as of 15:11, 13 February 2021

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?

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

Video Solution by hippopotamus1:

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

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

Solution 2

$x^{2^{\sqrt{2}}} = {\sqrt{2}}^{2^x}$

$2^{\sqrt{2}} \log x = 2^{x} \log \sqrt{2}$ (At this point we see by inspection that $x=\sqrt{2}$ is a solution.)

$\sqrt{2} \log 2 + \log \log x = x \log 2 + \log \log \sqrt{2}$

$\sqrt{2} + \log_2 \log_2 x = x + \log_2 \log_2 \sqrt{2} = x -1.$

$\log_2 \log_2 x = x - 1 - \sqrt{2}.$

LHS is a line. RHS 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 $\boxed{2 \leq S < 6}.$

~ ccx09

Video Solution by OmegaLearn (Logarithmic Tricks)

https://youtu.be/uCTpLB-kGR4

~ pi_is_3.14

@@ Line 15: / Line 15: @@
 == Solution 2 ==
 <math>x^{2^{\sqrt{2}}} = {\sqrt{2}}^{2^x}</math>
 <math>2^{\sqrt{2}} \log x = 2^{x} \log \sqrt{2}</math> (At this point we see by inspection that <math>x=\sqrt{2}</math> is a solution.)
 <math>\sqrt{2} \log 2 + \log \log x = x \log 2 + \log \log \sqrt{2}</math>
 <math>\sqrt{2} + \log_2 \log_2 x = x + \log_2 \log_2 \sqrt{2} = x -1.</math>
 <math>\log_2 \log_2 x = x - 1 - \sqrt{2}.</math>

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