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


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I don't understand how this answer could be downvoted. – Hakim Jun 24 '14 at 20:00
@Hakim ,I ask about the R.H. :4 ,I meant where is 4 ? – zeraoulia rafik Oct 4 '15 at 14:31
@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? – Did Nov 6 '15 at 19:10


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

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Thank you, this was really beautiful to discover :) – Kothas Jun 24 '14 at 13:28
Beauty is a major part of mathematics. Please trust the old man ! Cheers :) – Claude Leibovici Jun 24 '14 at 13:33
But is $x=2$ a solution? This proves only that no $x\ne2$ is solution. – Did Jun 24 '14 at 13:53
@Did One would only need to mention that the "iterated thing" converges. – Hakim Nov 6 '15 at 17:07
@Hakim "Mention"? No, prove! – Did Nov 6 '15 at 17:13

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


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

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It has been said, explicitely in a comment and implicitely in an answer. – Did Jun 24 '14 at 13:55
I didn't think of it this way. Nice. – Kye W Shi Jun 24 '14 at 14:07
@Did Hmm... Guess I didn't load new answers and comments. There were 2 answers when I started typing my answer. – Mike Jun 24 '14 at 14:34

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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Elegant. I'm a little confused on how sqrt(2x) = 2 became 2|x| = sqrt(2), though. – Kye W Shi Jun 24 '14 at 13:49
Oh, 2|x| = 4 makes sense now. – Kye W Shi Jun 24 '14 at 14:02
As commented on another answer: "But is $x=2$ a solution? This proves only that no $x≠2$ is solution." – Did Oct 4 '15 at 14:16

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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