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

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m (Solution 3)
 
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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 1 ==
 +
Note that
 +
<cmath>\begin{align*}
 +
x^{2^{\sqrt{2}}} &= {\sqrt{2}}^{2^x} \\
 +
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.)
  
Video solution by hippopotamus1:
+
We simplify the RHS, then take the base-<math>2</math> logarithm for both sides:
https://www.youtube.com/watch?v=GjO6C_qC13U&feature=youtu.be
+
<cmath>\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*}</cmath>
 +
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>
 +
 
 +
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>
 +
The answer is <math>\boxed{\textbf{(D) }2\le S<6}.</math>
 +
 
 +
~ccx09
 +
 
 +
==Solution 2==
 +
We rewrite the right side without using square roots, then take the base-<math>2</math> logarithm for both sides:
 +
<cmath>\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*}</cmath>
 +
By observations, <math>x=\sqrt2</math> is one solution. Graphing <math>f(x)=2^{\sqrt2}\log_2{x}</math> and <math>g(x)=2^{x-1},</math> we conclude that <math>(*)</math> has two solutions, with the smaller solution <math>x=\sqrt2.</math> We construct the following table of values:  
 +
<cmath>\begin{array}{c|c|c|c}
 +
& & & \\ [-2ex]
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\boldsymbol{x} & \boldsymbol{f(x)=2^{\sqrt2}\log_2{x}} & \boldsymbol{g(x)=2^{x-1}} & \textbf{Comparison} \\ [1ex]
 +
\hline
 +
& & & \\ [-1ex]
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1 & 0 & 1 & \\ [1ex]
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\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]
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\end{array}</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>
 +
Finally, approximating with <math>\sqrt2\approx1.414</math> results in <math>\boxed{\textbf{(D) }2\le S<6}.</math>
 +
 
 +
The graphs of <math>y=f(x)</math> and <math>y=g(x)</math> are shown below:
 +
<asy>
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/* Made by MRENTHUSIASM */
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size(1200,200);
 +
 
 +
int xMin = 0;
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int xMax = 5;
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int yMin = 0;
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int yMax = 5;
 +
 
 +
//Draws the horizontal gridlines
 +
void horizontalLines()
 +
{
 +
  for (int i = yMin+1; i < yMax; ++i)
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  {
 +
    draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
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  }
 +
}
 +
 
 +
//Draws the vertical gridlines
 +
void verticalLines()
 +
{
 +
  for (int i = xMin+1; i < xMax; ++i)
 +
  {
 +
    draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
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  }
 +
}
 +
 
 +
//Draws the horizontal ticks
 +
void horizontalTicks()
 +
{
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  for (int i = yMin+1; i < yMax; ++i)
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  {
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    draw((-1/8,i)--(1/8,i), black+linewidth(1));
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  }
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}
 +
 
 +
//Draws the vertical ticks
 +
void verticalTicks()
 +
{
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  for (int i = xMin+1; i < xMax; ++i)
 +
  {
 +
    draw((i,-1/8)--(i,1/8), black+linewidth(1));
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  }
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}
 +
 
 +
horizontalLines();
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verticalLines();
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horizontalTicks();
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verticalTicks();
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draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
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draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
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label("$x$",(xMax,0),(2,0));
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label("$y$",(0,yMax),(0,2));
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real f(real x) {return 2^sqrt(2)*log(x)/log(2);};
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real g(real x) {return 2^(x-1);};
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draw(graph(f,1,3.65),red,"$y=2^{\sqrt2}\log_2{x}$");
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draw(graph(g,0,3.32),blue,"$y=2^{x-1}$");
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pair A, B;
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A = intersectionpoint(graph(f,1,2),graph(g,1,2));
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B = intersectionpoint(graph(f,2,4),graph(g,2,4));
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dot(A,linewidth(4.5));
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dot(B,linewidth(4.5));
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label("$0$",(0,0),2.5*SW);
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label("$\sqrt2$",(A.x,0),2.25*S);
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label("$t$",(B.x,0),3*S);
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label("$4$",(4,0),3*S);
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label("$4$",(0,4),3*W);
  
== Solution (Rough Approximation) ==
+
draw(A--(A.x,0),dashed);
Note that this solution is not recommended unless you're running out of time.
+
draw(B--(B.x,0),dashed);
  
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
+
add(legend(),point(E),40E,UnFill);
 +
</asy>
 +
~MRENTHUSIASM
  
 
== Video Solution by OmegaLearn (Logarithmic Tricks) ==
 
== Video Solution by OmegaLearn (Logarithmic Tricks) ==
Line 18: Line 135:
 
~ pi_is_3.14
 
~ pi_is_3.14
  
 +
==Video Solution by hippopotamus1==
 +
https://www.youtube.com/watch?v=GjO6C_qC13U&feature=youtu.be
 +
 +
==Video Solution by The Power of Logic==
 +
https://youtu.be/0i6qoGpk_Ew
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|num-b=20|num-a=22}}
 
{{AMC12 box|year=2021|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:22, 24 October 2022

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

We rewrite the right side without using square roots, 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. Graphing $f(x)=2^{\sqrt2}\log_2{x}$ and $g(x)=2^{x-1},$ we conclude that $(*)$ has two solutions, with the smaller solution $x=\sqrt2.$ 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 $\boxed{\textbf{(D) }2\le S<6}.$

The graphs of $y=f(x)$ and $y=g(x)$ are shown below: [asy] /* Made by MRENTHUSIASM */ size(1200,200);   int xMin = 0; int xMax = 5; int yMin = 0; int yMax = 5;  //Draws the horizontal gridlines void horizontalLines() {   for (int i = yMin+1; i < yMax; ++i)   {     draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));   } }  //Draws the vertical gridlines void verticalLines() {   for (int i = xMin+1; i < xMax; ++i)   {     draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));   } }  //Draws the horizontal ticks void horizontalTicks() {   for (int i = yMin+1; i < yMax; ++i)   {     draw((-1/8,i)--(1/8,i), black+linewidth(1));   } }  //Draws the vertical ticks void verticalTicks() {   for (int i = xMin+1; i < xMax; ++i)   {     draw((i,-1/8)--(i,1/8), black+linewidth(1));   } }  horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2));  real f(real x) {return 2^sqrt(2)*log(x)/log(2);}; real g(real x) {return 2^(x-1);};  draw(graph(f,1,3.65),red,"$y=2^{\sqrt2}\log_2{x}$"); draw(graph(g,0,3.32),blue,"$y=2^{x-1}$");  pair A, B; A = intersectionpoint(graph(f,1,2),graph(g,1,2)); B = intersectionpoint(graph(f,2,4),graph(g,2,4)); dot(A,linewidth(4.5)); dot(B,linewidth(4.5));  label("$0$",(0,0),2.5*SW); label("$\sqrt2$",(A.x,0),2.25*S); label("$t$",(B.x,0),3*S); label("$4$",(4,0),3*S); label("$4$",(0,4),3*W);  draw(A--(A.x,0),dashed); draw(B--(B.x,0),dashed);  add(legend(),point(E),40E,UnFill); [/asy] ~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

Video Solution by The Power of Logic

https://youtu.be/0i6qoGpk_Ew

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

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

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