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Today I faced a strange equation and I didn't manage to find a solution to it: $$x\sqrt{x\sqrt{x\sqrt{x\dots}}} = 4$$ Maybe someone will help me to find a way to solve it. By the way, this equation is from high school course.

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  • $\begingroup$ What does this have to do with geometric progressions? $\endgroup$ – Gerry Myerson Jun 24 '14 at 13:17
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    $\begingroup$ Well this equation was in geometric progressions chapter in my book. $\endgroup$ – Kothas Jun 24 '14 at 13:28
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    $\begingroup$ The exponents form a geometric progression (first term 1, ratio 1/2). $\endgroup$ – Did Jun 24 '14 at 13:52
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    $\begingroup$ @GerryMyerson It has everything to do with geometric progressions. :) $\endgroup$ – Akiva Weinberger Oct 4 '15 at 14:22
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$$x\sqrt{x\sqrt{x\sqrt{x\cdots}}}=x\,x^{1/2}\,x^{1/4}\,x^{1/8}\cdots=x^{1+1/2+1/4+1/8+\cdots}=x^2$$

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    $\begingroup$ I don't understand how this answer could be downvoted. $\endgroup$ – Hakim Jun 24 '14 at 20:00
  • $\begingroup$ @Hakim ,I ask about the R.H. :4 ,I meant where is 4 ? $\endgroup$ – zeraoulia rafik Oct 4 '15 at 14:31
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    $\begingroup$ @zeraouliarafik I just realized that your comment could be read as a justification for having downvoted my answer. Is it? And you downvoted this because there is no 4 in it? Seriously? $\endgroup$ – Did Nov 6 '15 at 19:10
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Hint

Divide both sides by $x$ and square them. You should notice something beautiful.

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  • $\begingroup$ Thank you, this was really beautiful to discover :) $\endgroup$ – Kothas Jun 24 '14 at 13:28
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    $\begingroup$ Beauty is a major part of mathematics. Please trust the old man ! Cheers :) $\endgroup$ – Claude Leibovici Jun 24 '14 at 13:33
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    $\begingroup$ But is $x=2$ a solution? This proves only that no $x\ne2$ is solution. $\endgroup$ – Did Jun 24 '14 at 13:53
  • $\begingroup$ @Did One would only need to mention that the "iterated thing" converges. $\endgroup$ – Hakim Nov 6 '15 at 17:07
  • $\begingroup$ @Hakim "Mention"? No, prove! $\endgroup$ – Did Nov 6 '15 at 17:13
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An alternative way $$x\sqrt{x\sqrt{x\sqrt{x\cdots}}}=4\implies \sqrt{x\sqrt{x\sqrt{x\sqrt{x\cdots}}}}=2.$$ Now observe that $$ \sqrt{x\underbrace{\sqrt{x\sqrt{x\sqrt{x\cdots}}}}_{2}}=2.$$ Therefore your equation reduces to $$\sqrt{2x}=2\implies 2x=4\implies x=2.\tag{$x>0$}$$

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  • $\begingroup$ Elegant. I'm a little confused on how sqrt(2x) = 2 became 2|x| = sqrt(2), though. $\endgroup$ – Kye W Shi Jun 24 '14 at 13:49
  • $\begingroup$ Oh, 2|x| = 4 makes sense now. $\endgroup$ – Kye W Shi Jun 24 '14 at 14:02
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    $\begingroup$ As commented on another answer: "But is $x=2$ a solution? This proves only that no $x≠2$ is solution." $\endgroup$ – Did Oct 4 '15 at 14:16
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Well, since it hasn't been said yet and you mentioned it was in the geometric progressions chapter of your book, note that the equation can be rewritten as

$$x^{1+1/2+1/4+1/8+...}=4$$

So there's your geometric progression. Note that $x$ should be positive.

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  • $\begingroup$ It has been said, explicitely in a comment and implicitely in an answer. $\endgroup$ – Did Jun 24 '14 at 13:55
  • $\begingroup$ I didn't think of it this way. Nice. $\endgroup$ – Kye W Shi Jun 24 '14 at 14:07
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    $\begingroup$ @Did Hmm... Guess I didn't load new answers and comments. There were 2 answers when I started typing my answer. $\endgroup$ – Mike Jun 24 '14 at 14:34
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A general approach to solve such problems is to exploit self similarity. In this particular problem, the term within the first square root is just the left hand side itself. Thus

$$ x\sqrt{4}=4. $$

As pointed out by Did in the comments to other questions, this shows that if a solution exists, then it is unique and equals $2$.

To show that $2$ is indeed a solution, you need to go the route via geometric series. (As an aside, geometric series themselves can be solved via self similarity: $ s:=1+q+q^2+\dots \Rightarrow s=1+qs $)

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The power of given variabe is in the term of sum of G.P . The indices of x is such that 1+1/2+1/4+...... Because we know that the infinite sum of G.P: Sn = a/(1-r) where r=common ratio

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