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) == | + | == 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. | ||
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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 | ||
Line 44: | Line 45: | ||
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) | + | 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 <math>t\in(2,4).</math> Finally, we conclude that <cmath>S=\sqrt2+t\in\left(\sqrt2+2,\sqrt2+4\right) | + | 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> Finally, we conclude that <cmath>S=\sqrt2+t\in\left(\sqrt2+2,\sqrt2+4\right).</cmath> |
− | + | Approximating <math>\sqrt2\approx1.414,</math> we get the answer <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/xyxzsvjort | Graphs of <math>f(x)</math> and <math>g(x)</math> in Desmos: https://www.desmos.com/calculator/xyxzsvjort |
Revision as of 19:11, 22 March 2021
Contents
Problem
Let be the sum of all positive real numbers
for which
Which of the following statements is true?
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 . From this, we can deduce that this equality has two solutions, since
grows faster than
(for greater values of
) and
is greater than
for
and less than
for
, where
is the second solution. Thus, the answer cannot be
or
. We then start plugging in numbers to roughly approximate the answer. When
,
, thus the answer cannot be
. Then, when
,
. Therefore,
, so the answer is
. ~Baolan
Solution 2
(At this point we see by inspection that
is a solution.)
RHS is a line. 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 3 (Cleaner Solution by Graphing and Very Light Approximations)
We rewrite the right side, then take the base- logarithm for both sides:
By observations,
is one solution. By quick sketches of
and
we know 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
Finally, we conclude that
Approximating
we get the answer
Graphs of and
in Desmos: https://www.desmos.com/calculator/xyxzsvjort
~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
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.