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.
 A: 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 $$\sqrt{2x}=2\implies 2x=4\implies x=2.\tag{$x>0$}$$
A: 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.
A: $$x\sqrt{x\sqrt{x\sqrt{x\cdots}}}=x\,x^{1/2}\,x^{1/4}\,x^{1/8}\cdots=x^{1+1/2+1/4+1/8+\cdots}=x^2$$
A: Hint
Divide both sides by $x$ and square them. You should notice something beautiful.
A: 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
$)
A: 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 
