This is a self-answer question taken from this youtube video. The youtube video describes the presented problem as a no calculator problem presented in 2021, as an Olympiad qualifying problem given in America. The youtube video cited AMC12B, problem 12.
$\underline{\text{The Problem}}$
Let $~P~$ denote the set of all positive (real number) solutions to
$\displaystyle x^{\left[ ~2^{\sqrt{2}} ~\right]} = \sqrt{2}^{\left[ ~2^x ~\right]}.$
Let $~S~$ denote the sum of all elements in $~P.$ Then, which of the following is true:
- A : $~\displaystyle S < \sqrt{2}.$
- B : $~\displaystyle S = \sqrt{2}.$
- C : $~\displaystyle \sqrt{2} < S < 2.$
- D : $~2 \leq S < 6.$
- E : $~6 \leq S.$
$\underline{\text{My Reaction To The Youtube Solution}}$
I only watched part of it, and then turned away. While I found the youtube solution creative, there were aspects of it that I didn't like. For one think, it applied nested logarithms (i.e. $~\log\{ ~\log[\text{expression}] ~\} ~$). Also, it involved more creativity than the problem required. For example, it examined logarithms, base 2.
Further, it didn't immediately focus on derivatives, which I regard as the primary weapon against this type of problem.
So, I have provided my own answer.
$~\underline{\text{Overview And Assignment of Functions}}$
Throughout this answer, $~\log(\text{expression})~$ refers to natural logarithms.
Throughout this answer, unless otherwise specified, the domain of any of the functions discussed is $~\Bbb{R^+}.$
This answer requires a basic knowledge of derivatives to understand.
Let:
$\displaystyle A(x) = x^{\left[ ~2^{\sqrt{2}} ~\right]} \implies $
$\displaystyle \log[A(x)] = 2^{\sqrt{2}} \times \log(x).$
$~\displaystyle B(x) = \sqrt{2}^{\left[ ~2^x ~\right]} \implies $
$\displaystyle \log[B(x)] = 2^x \times \log{\sqrt{2}} = \frac{1}{2} \times 2^x \times \log(2).$
$f(x) = A(x) - B(x).$
So, $~x \in P \iff \{ ~[x > 0] \wedge [f(x) = 0] ~\}.$
$g(x) = \log[A(x)] - \log[B(x)].$
Since the function $~y = \log(x)~$ is strictly increasing on $~\Bbb{R^+},~$ you have that:
$$f(x) = 0 \iff A(x) = B(x) \iff \log[A(x)] = \log[B(x)] \iff g(x) = 0.$$
One of the points brought out in the Youtube video is that you can tell by inspection that $~\displaystyle f\left[ ~\sqrt{2} ~\right] = 0.~$ While this is true, this type of creativity is not part of my standard attack plan for this type of problem.
Given the multiple choice options, and the prohibition against calculators, my approach will be to examine, by inspection, $~g(x), g'(x), ~$ and (by inference) $~g''(x),~$ for small positive integers, where the examination does not require a calculator.
Notes:
One thing that the Youtube video did, which I absolutely agree with, is (in effect) to compute the logarithms of $~A(x)~$ and $~B(x).~$ For this type of problem, with nested exponentials on both sides, I think that this is generally a good idea to (at least) attempt.
$\displaystyle \log[f(x)] = \log[A(x) - B(x)] = \log\left[ ~\dfrac{A(x)}{B(x)} ~\right] \neq \log[A(x)] - \log[B(x)] = g(x).~$
I mention this to avoid confusion with the following point.
When $~x_0~$ is determined such that $~g'(x_0) = 0,~$ this will represent that $~g(x)~$ is maximized at $~(x = x_0).~$ It is valid to then use this result to identify the regions where the roots of $~g(x) = 0~$ exist. Since these roots of $~g(x)~$ also represent the roots of $~f(x) = 0,~$ this solves the problem.
However, it would be invalid to then conclude (necessarily) that $~f(x)~$ attains a maximum value at $~(x=x_0),~$ since $~g'(x) = 0~$ does not (to the best of my knowledge) imply that $~f'(x) = 0.$
Fortunately, the problem, which does not ask for the maximum value of $~f(x),~$ does not require any coordination between $~f'(x)~$ and $~g'(x).$
$~\underline{\text{Inspection of} ~g(x)}$
$~x \leq 1.$
This implies that $~\log(x) \leq 0 \implies \log[A(x)] \leq 0.~$
Since $~\log[B(x)]~$ is always positive, you have that $~g(x) < 0 ~: ~x \leq 1.$
$~x = 2.$
$\displaystyle g(2) = \left[ ~2^{\sqrt{2}} \times \log(2) ~\right] - \log(2) > 0.$
$~x = 4.$
$\displaystyle g(4) = 2^{\sqrt{2}} \times 2\log(2) - 8\log(2).$
Since $~\displaystyle \sqrt{2} < 2, ~2^{\sqrt{2}} < 2^2 = 4.~$ Therefore $~g(4) < 0.~$
$~\underline{\text{Inspection of} ~g'(x)}$
$\displaystyle A'(x) = \frac{2^{\sqrt{2}}}{x}.$
$\displaystyle B'(x) = \frac{1}{2} \times [\log(2)]^2 \times 2^x.$
$~g'(x) = A'(x) - B'(x).$
$\displaystyle 2 < e < 2^2 \implies \frac{1}{2} < \log(2) < 1.$
So:
$\displaystyle A'(1) = 2^{\sqrt{2}}, ~B'(1) = \left[ ~\log(2) ~\right]^2 \implies g'(1) > 0.$
As $~x~$ decreases, from $~1~$ towards $~0,~$
$\displaystyle~\frac{1}{x}~$ increases and $~2^x~$ decreases.
This implies that as $~x~$ decreases, from $~1~$ towards $~0,~$ $~\displaystyle ~A'(x)~$ increases and $~B'(x)~$ decreases.
Therefore, for all $~x~$ such that $~0 < x < 1,~$ you have that $~g'(x) = A'(x) - B'(x) > 0.~$
Similarly, as $~x~$ increases from $~1, ~A(x)~$ decreases and $~B(x)~$ increases.
Therefore, for all $~x~$ such that $~1 < x,~$
you have that $~g'(x)~$ is a decreasing function.
Based on the previous bullet point, for all $~x > 1,~$
you have that $~g''(x) < 0.~$
$\displaystyle A'(4) = \frac{2^{\sqrt{2}}}{4} < 1.$ $~\displaystyle B'(4) = 8\left[ ~\log(2) ~\right]^2 > 8[0.5]^2 = 2.$
Therefore $~g'(4) < 0.$
$~\underline{\text{Final Analysis}}$
$~g'(x),~$ is positive at $~(x \leq 1),~$ negative at $~(x=4),~$ and is strictly decreasing for $~x > 1.~$ Consequently, since $~g'(x)~$ is a continuous function, there must exist a unique point $~(x = x_0)~$ such that $~g'(x_0) = 0.~$
Further, you must have that $~1 < x_0 < 4,~$ and that $~g''(x_0) < 0.$
Therefore, $~g(x)~$ attains a maximum value at $~x = x_0.~$
You also have that $~g(1) < 0, ~g(2) > 0,~$ and $~g(4) < 0.$
Therefore, $~g(x_0) > 0,~$ there must be exactly two roots of $~g(x) = 0,~$ and both of these roots must lie in the interval $~(1,4).$
Further, regardless of whether $~2 < x_0, ~2 = x_0, ~$ or $~2 > x_0,~$ one of the two roots must be $~< 2,~$ and the other root must be $~> 2.$
Also, as discussed, the roots of $~f(x)~$ must coincide with the roots of $~g(x).$
Option E is therefore false, so Option D must be the right answer.