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

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<math>\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</math>
 
<math>\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</math>
  
== Solution (Rough Approximation) ==
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== Solution 1 (Rough Approximation) ==
 
Note that this solution is not recommended unless you're running out of time.
 
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 <math>x=\sqrt{2}</math>. From this, we can deduce that this equality has two solutions, since <math>\sqrt{2}^{2^{x}}</math> grows faster than <math>x^{2^{\sqrt{2}}}</math> (for greater values of <math>x</math>) and <math>\sqrt{2}^{2^{x}}</math> is greater than <math>x^{2^{\sqrt{2}}}</math> for <math>x<\sqrt{2}</math> and less than <math>x^{2^{\sqrt{2}}}</math> for <math>\sqrt{2}<x<n</math>, where <math>n</math> is the second solution. Thus, the answer cannot be <math>\text{A}</math> or <math>\text{B}</math>. We then start plugging in numbers to roughly approximate the answer. When <math>x=2</math>, <math>x^{2^{\sqrt{2}}}>\sqrt{2}^{2^{x}}</math>, thus the answer cannot be <math>\text{C}</math>. Then, when <math>x=4</math>, <math>x^{2^{\sqrt{2}}}=4^{2^{\sqrt{2}}}<64<\sqrt{2}^{2^{x}}=256</math>. Therefore, <math>S<4+\sqrt{2}<6</math>, so the answer is <math>\boxed{\textbf{(D) } 2 \le S < 6}</math>. ~Baolan
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Upon pure observation, it is obvious that one solution to this equality is <math>x=\sqrt{2}</math>. From this, we can deduce that this equality has two solutions, since <math>\sqrt{2}^{2^{x}}</math> grows faster than <math>x^{2^{\sqrt{2}}}</math> (for greater values of <math>x</math>) and <math>\sqrt{2}^{2^{x}}</math> is greater than <math>x^{2^{\sqrt{2}}}</math> for <math>x<\sqrt{2}</math> and less than <math>x^{2^{\sqrt{2}}}</math> for <math>\sqrt{2}<x<n</math>, where <math>n</math> is the second solution. Thus, the answer cannot be <math>\text{A}</math> or <math>\text{B}</math>. We then start plugging in numbers to roughly approximate the answer. When <math>x=2</math>, <math>x^{2^{\sqrt{2}}}>\sqrt{2}^{2^{x}}</math>, thus the answer cannot be <math>\text{C}</math>. Then, when <math>x=4</math>, <math>x^{2^{\sqrt{2}}}=4^{2^{\sqrt{2}}}<64<\sqrt{2}^{2^{x}}=256</math>. Therefore, <math>S<4+\sqrt{2}<6</math>, so the answer is <math>\boxed{\textbf{(D) }2\le S<6}</math>.
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~Baolan
  
 
== Solution 2 ==
 
== Solution 2 ==
<math>x^{2^{\sqrt{2}}} = {\sqrt{2}}^{2^x}</math>
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Note that
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<cmath>\begin{align*}
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x^{2^{\sqrt{2}}} &= {\sqrt{2}}^{2^x} \\
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2^{\sqrt{2}} \log_2 x &= 2^{x} \log_2 \sqrt{2}.
 +
\end{align*}</cmath>
 +
(At this point we see by inspection that <math>x=\sqrt{2}</math> is a solution.)
  
<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.)
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We simplify the RHS, then take the base-<math>2</math> logarithm for both sides:
 
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<cmath>\begin{align*}
<math>\sqrt{2} \log 2 + \log \log x = x \log 2 + \log \log \sqrt{2}</math>
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2^{\sqrt{2}} \log_2 x &= 2^{x-1} \\
 
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\log_2{\left(2^{\sqrt{2}} \log_2 x\right)} &= x-1 \\
<math>\sqrt{2} + \log_2 \log_2 x = x + \log_2 \log_2 \sqrt{2} = x -1.</math>
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\sqrt{2} + \log_2 \log_2 x &= x-1 \\
 
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\log_2 \log_2 x &= x - 1 - \sqrt{2}.
<math>\log_2 \log_2 x = x - 1 - \sqrt{2}.</math>
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\end{align*}</cmath>
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The RHS is a line; the LHS is a concave curve that looks like a logarithm and has <math>x</math> intercept at <math>(2,0).</math>
  
RHS is a line. LHS is a concave curve that looks like a logarithm and has <math>x</math> intercept at <math>(2,0).</math>
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There are at most two solutions, one of which is <math>\sqrt{2}.</math> But note that at <math>x=2,</math> we have <math>\log_2 \log_2 {2} = 0 > 2 - 1 - \sqrt{2},</math> meaning that the log log curve is above the line, so it must intersect the line again at a point <math>x > 2.</math> Now we check <math>x=4</math> and see that <math>\log_2 \log_2 {4} = 1 < 4 - 1 - \sqrt{2},</math> which means at <math>x=4</math> the line is already above the log log curve. Thus, the second solution lies in the interval <math>(2,4).</math>
There are at most two solutions, one of which is <math>\sqrt{2}.</math> But note that at <math>x=2,</math> we have <math>\log_2 \log_2 (2) = 0 > 2 - 1 - \sqrt{2},</math> meaning that the log log curve is above the line, so it must intersect the line again at a point <math>x > 2.</math> Now we check <math>x=4</math> and see that <math>\log_2 \log_2 (4) = 1 < 4 - 1 - \sqrt{2},</math> which means at <math>x=4</math> the line is already above the log log curve. Thus, the second solution lies in the interval <math>(2,4).</math>
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The answer is <math>\boxed{\textbf{(D) }2\le S<6}.</math>
The answer is <math>\boxed{2 \leq S < 6}.</math>
 
  
 
~ ccx09
 
~ ccx09
  
==Solution 3 (Cleaner Solution by Graphing and Very Light Approximations)==
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==Solution 3 (Graphs and Light Approximations)==
 
We rewrite the right side, then take the base-<math>2</math> logarithm for both sides:
 
We rewrite the right side, then take the base-<math>2</math> logarithm for both sides:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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x^{2^{\sqrt2}}&=2^{2^{x-1}} \\
 
x^{2^{\sqrt2}}&=2^{2^{x-1}} \\
 
\log_2{\left(x^{2^{\sqrt2}}\right)}&=\log_2{\left(2^{2^{x-1}}\right)} \\
 
\log_2{\left(x^{2^{\sqrt2}}\right)}&=\log_2{\left(2^{2^{x-1}}\right)} \\
2^{\sqrt2}\log_2{x}&=2^{x-1}. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)
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2^{\sqrt2}\log_2{x}&=2^{x-1}. \hspace{20mm} (*)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
By observations, <math>x=\sqrt2</math> is one solution. By quick sketches of <math>f(x)=2^{\sqrt2}\log_2{x}</math> and <math>g(x)=2^{x-1},</math> we know that <math>(*)</math> has two solutions, with <math>x=\sqrt2</math> the smaller solution. We construct the following table of values:  
 
By observations, <math>x=\sqrt2</math> is one solution. By quick sketches of <math>f(x)=2^{\sqrt2}\log_2{x}</math> and <math>g(x)=2^{x-1},</math> we know that <math>(*)</math> has two solutions, with <math>x=\sqrt2</math> the smaller solution. We construct the following table of values:  
 
<cmath>\begin{array}{c|c|c|c}
 
<cmath>\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]
 
\boldsymbol{x} & \boldsymbol{f(x)=2^{\sqrt2}\log_2{x}} & \boldsymbol{g(x)=2^{x-1}} & \textbf{Comparison} \\ [1ex]
 
\hline
 
\hline
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2 & 2^{\sqrt2} & 2 & f(2)>g(2) \\ [1ex]
 
2 & 2^{\sqrt2} & 2 & f(2)>g(2) \\ [1ex]
 
3 & 2^{\sqrt2}\log_2{3} & 2^2 & \\ [1ex]
 
3 & 2^{\sqrt2}\log_2{3} & 2^2 & \\ [1ex]
4 & 2^{\sqrt2+1} & 2^3 & f(4)<g(4)
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4 & 2^{\sqrt2+1} & 2^3 & f(4)<g(4) \\ [1ex]
 
\end{array}</cmath>
 
\end{array}</cmath>
Let <math>x=t</math> be the larger solution. Since exponential functions outgrow logarithmic functions, we have <math>f(x)<g(x)</math> for all <math>x>t.</math> By the <b>Intermediate Value Theorem</b>, we get that <math>t\in(2,4).</math> Finally, we conclude that <cmath>S=\sqrt2+t\in\left(\sqrt2+2,\sqrt2+4\right),</cmath>
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Let <math>x=t</math> be the larger solution. Since exponential functions outgrow logarithmic functions, we have <math>f(x)<g(x)</math> for all <math>x>t.</math> By the <b>Intermediate Value Theorem</b>, we get <math>t\in(2,4),</math> from which <cmath>S=\sqrt2+t\in\left(\sqrt2+2,\sqrt2+4\right).</cmath>
and the answer must be <math>\boxed{\textbf{(D) }2\le S<6}.</math>
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Finally, approximating with <math>\sqrt2\approx1.414</math> results in <math>\boxed{\textbf{(D) }2\le S<6}.</math>
  
Graphs of <math>f(x)</math> and <math>g(x)</math> in Desmos: https://www.desmos.com/calculator/o24gbnesq3
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Graphs of <math>f(x)</math> and <math>g(x)</math> in Desmos: https://www.desmos.com/calculator/xyxzsvjort
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
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~ pi_is_3.14
 
~ pi_is_3.14
  
==Video Solution by hippopotamus1:==
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==Video Solution by hippopotamus1==
 
https://www.youtube.com/watch?v=GjO6C_qC13U&feature=youtu.be
 
https://www.youtube.com/watch?v=GjO6C_qC13U&feature=youtu.be
  

Latest revision as of 04:38, 25 August 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?

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

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 $\sqrt{2}<x<n$, 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

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: \[\begin{array}{c|c|c|c} & & & \\ [-2ex] \boldsymbol{x} & \boldsymbol{f(x)=2^{\sqrt2}\log_2{x}} & \boldsymbol{g(x)=2^{x-1}} & \textbf{Comparison} \\ [1ex] \hline & & & \\ [-1ex] 1 & 0 & 1 & \\ [1ex] \sqrt2 & 2^{\sqrt2-1} & 2^{\sqrt2-1} & f\left(\sqrt2\right)=g\left(\sqrt2\right) \\ [1ex] 2 & 2^{\sqrt2} & 2 & f(2)>g(2) \\ [1ex] 3 & 2^{\sqrt2}\log_2{3} & 2^2 & \\ [1ex] 4 & 2^{\sqrt2+1} & 2^3 & f(4)<g(4) \\ [1ex] \end{array}\] 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

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

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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