Difference between revisions of "2021 AMC 12B Problems/Problem 21"
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2^{\sqrt2}\log_2{x}&=2^{x-1}. \hspace{20mm} (*) | 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. 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 <math>x=\sqrt2</math> | + | 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} | <cmath>\begin{array}{c|c|c|c} | ||
& & & \\ [-2ex] | & & & \\ [-2ex] |
Revision as of 19:10, 28 November 2021
Contents
Problem
Let be the sum of all positive real numbers
for which
Which 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
Solution 3
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
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.