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 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.) | ||
− | = | + | We simplify the RHS, then take the base-<math>2</math> logarithm for both sides: |
− | + | <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] | ||
+ | \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}</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 <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> | ||
+ | /* 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) == | == Video Solution by OmegaLearn (Logarithmic Tricks) == | ||
Line 15: | 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
Contents
Problem
Let be the sum of all positive real numbers for whichWhich of the following statements is true?
Solution 1
Note that (At this point we see by inspection that is a solution.)
We simplify the RHS, then take the base- logarithm for both sides: The RHS is a line; the LHS is a concave curve that looks like a logarithm and has intercept at
There are at most two solutions, one of which is But note that at we have meaning that the log log curve is above the line, so it must intersect the line again at a point Now we check and see that which means at the line is already above the log log curve. Thus, the second solution lies in the interval The answer is
~ccx09
Solution 2
We rewrite the right side without using square roots, then take the base- logarithm for both sides: By observations, is one solution. Graphing and we conclude that has two solutions, with the smaller solution We construct the following table of values: Let be the larger solution. Since exponential functions outgrow logarithmic functions, we have for all By the Intermediate Value Theorem, we get from which Finally, approximating with results in
The graphs of and are shown below: ~MRENTHUSIASM
Video Solution by OmegaLearn (Logarithmic Tricks)
~ 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
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
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.