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I tried to prove that $2$ is the only solution to the equation $x=\sqrt{2}^x$ without any results.

Here's my try : Let $f:[0,+\infty)\rightarrow\mathbb{R}$ such that $f(x)=\sqrt{2}^x-x$. Thus, $f'(x)=\sqrt{2}^x\ln\sqrt2-1$. $f'$ converges to $0$ when $x=\log_{\sqrt2}\frac{1}{\ln\sqrt2}$. The derivate should keep it's sign to prove that $f$ has exactly one solution, I can't understand what is going on. Please, help

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    $\begingroup$ It's not. $4 = \sqrt{2^4}$ $\endgroup$
    – fleablood
    Oct 1 '16 at 20:26
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It is not the only solution: $x=4$ is another solution. Actually, you can show, in your notations, that

  • $f'(x)<0\;$ for $\;0\le x<\dfrac{2(\ln 2-\ln(\ln 2))}{\ln2}\approx 3.06$,
  • $f'(x)>0\;$ for $\;x> \dfrac{2(\ln 2-\ln(\ln 2))}{\ln2}$.

So there are only two solutions: $2$ and $4$.

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  • $\begingroup$ I think the critical point (which is a minimum) is actually at $x = -2 \dfrac{\ln\ln\sqrt{2}}{\ln 2}$, which is slightly more than $3$, which makes sense, since it should lie between the two zeroes at $x=2$ and $x=4$. $\endgroup$
    – tracing
    Oct 2 '16 at 5:14
  • $\begingroup$ Why its $-\frac{2\ln(\ln2)}{\ln2}$ and not $\log_{\sqrt2}\frac{1}{\ln\sqrt2}$ ? $\endgroup$
    – user180321
    Oct 2 '16 at 9:06
  • $\begingroup$ @user180321: because I wrote ${\sqrt2} ^x$ as $2^{\tfrac x2}$. $\endgroup$
    – Bernard
    Oct 2 '16 at 9:49
  • $\begingroup$ @tracing: You're right, I had forgotten a 2 before the fraction. I've updated my answer. Thank you for pointing it! $\endgroup$
    – Bernard
    Oct 2 '16 at 9:51
  • $\begingroup$ @Bernard well, actually $2-\frac{2\ln(\ln2)}{\ln2}\approx 2.52$ and NOT close to $3.06$ . I also checked from my calculator that $f'\left(2-\frac{2\ln(\ln2)}{\ln2}\right)\approx -1.7$. Can you explain how did you find these inverses ? Thank you- $\endgroup$
    – user180321
    Oct 2 '16 at 17:58
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Looks like $2$ isn't the only solution. If you take the power $\frac{1}{x}$ to each side you get

$$x^{\frac{1}{X}} = 2^{\frac{1}{2}}$$

Obviously, $x=2$ works, but also notice that

$$4^{\frac{1}{4}} = (2^2)^{\frac{1}{4}} = 2^{\frac{1}{2}}$$

So $x=4$ is also a solution.

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The function $f(x)=\frac {1}{x}\ln x$ is strictly increasing for $x\in [1,e]$ and strictly decreasing for $x\geq e$ because $f'(x)=\frac {1-\ln x}{x^2}.$ Also $f(1)=0=\lim_{x\to \infty}f(x).$ So for any $u\in (0,e)$ there is a unique $v>e$ such that $f(u)=f(v).$ That is, $u^{1/u}=v^{1/v}.$ In particular when $u=2$ we have $v=4.$

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Let $f(x) = e^{\frac x 2\ln 2}$ and rewrite your equation to $f(x)=x$. If $x = 2$ were a unique solution, then the line $y= x$ would be tangent to graph of $f(x)$. But, $f'(x) = \frac{\ln 2}2e^{\frac x 2\ln 2}$, hence $f'(2) = \ln 2\neq 1$, which is slope of $y = x$. Thus, the solution is not unique.

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