# Solve $x\sqrt{x\sqrt{x\sqrt{x\dots}}} = 4$

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

$$x\sqrt{x\sqrt{x\sqrt{x\cdots}}}=x\,x^{1/2}\,x^{1/4}\,x^{1/8}\cdots=x^{1+1/2+1/4+1/8+\cdots}=x^2$$

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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

Hint

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

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

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

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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
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 $$\sqrt{2x}=2\implies 2x=4\implies x=2.\tag{x>0}$$
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