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.
2026-03-25 14:03:25.1774447405
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Solve $x\sqrt{x\sqrt{x\sqrt{x\dots}}} = 4$
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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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$$x\sqrt{x\sqrt{x\sqrt{x\cdots}}}=x\,x^{1/2}\,x^{1/4}\,x^{1/8}\cdots=x^{1+1/2+1/4+1/8+\cdots}=x^2$$
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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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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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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 $)
Hint
Divide both sides by $x$ and square them. You should notice something beautiful.