# Geometric progression equation with irrational numbers

today I faced quite 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 at 13:17
Well this equation was in geometric progressions chapter in my book. –  Kothas Jun 24 at 13:28
The exponents form a geometric progression (first term 1, ratio 1/2). –  Did Jun 24 at 13:52

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

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I don't understand how this answer could be downvoted. –  Hakim Jun 24 at 20:00

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 at 13:28
Beauty is a major part of mathematics. Please trust the old man ! Cheers :) –  Claude Leibovici Jun 24 at 13:33
But is $x=2$ a solution? This proves only that no $x\ne2$ is solution. –  Did Jun 24 at 13:53

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 at 13:55
I didn't think of it this way. Nice. –  Kye W Shi Jun 24 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: $$\sqrt{2x}=2\implies 2x=4\implies x=2.\tag{x>0}$$
Since squaring $\sqrt{2x}$ never yields $-2x$, I fail to see why the absolute value appears. –  Did Jun 24 at 14:11