Solving $3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{16} + ...}}}}$ 
How to find the value of $3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{16} + ...}}}}$

I tried to solve it and found a relation that if I assume the given expression to be something say $x$, then following result holds true.
$$\boxed{x = 3 + \sqrt{3x}}\hspace{4cm}\bf ...(1.)$$
Can be proved as,
$$x = 3 + \sqrt{3(3 + \sqrt{3x})}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^3x}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^3(3 + \sqrt{3x})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^7x})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^7(3 + \sqrt{3x})})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{15}x}}}}$$
If I continue this process repeatedly,
$$x= 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^{8} + \sqrt{3^{16} +\sqrt{...}}}}}$$
But how can I make sure that $(1.)$ is absolutely correct? Also how can I make sure that "If I continue this process repeatedly" I would get the same expression?

Let
$$f(x)=x+\sqrt{x^2+\sqrt{x^4+\sqrt{ x^8+\cdots}}},$$
which is to be understood as $\lim_{n\to\infty}f_n(x)$, where
$$f_n(x)=x+\sqrt{x^2+\sqrt{x^4+\sqrt{ x^8+\cdots\sqrt{\cdots\sqrt{x^{2^n}}}}}},$$
or recursively (and thus finally in a sufficiently formal way),
$$\tag1f_0(x)=x,\qquad f_{n+1}(x)=x+\sqrt{f_n(x^2)}.$$
We want to find the limit (if it exists) $f(3)$.
Note that instead of your equation, $(1)$ rather gives us
$$\tag2 {f(x)=x+\sqrt{f(x^2)}}.$$
Your guess seems to have ignored the way the exponents of $x$ grow under the deeper radicals.
The above looks like we need to first find $f$ evaluated at larger arguments (hence with seemingly "worse" convergence), but we can easily get out of that trap:

 For $x>0$, let $g_n(x)=\frac1xf_n(x)$. Then $(1)$ becomes $$ g_0(x)=1,\qquad g_{n+1}(x)=1+\frac1x\sqrt{f_n(x^2)}=1+\sqrt{\frac1{x^2}f_n(x^2)}=1+\sqrt{g_n(x^2)}.$$


 In particular, we find by induction that all $g_n$ are constant, $g_n(x)=G_n$. What about convergence of this sequence of constants $$\tag3 G_0=1,\qquad G_{n+1}=1+\sqrt{G_n}?$$


 As $x\mapsto 1+\sqrt x$ is an increasing function on $[0,\infty)$ and as $G_1=2>1=G_0$, we see that $\{G_n\}_n$ is inceasing. On the other hand, we readily see by induction that $G_n<3$. Hence $G:=\lim_{n\to\infty}G_n$ exists, and certainlky $1<G\le 3$.


 From $(2)$, we know that $G$ must be a soution of $$ x=1+\sqrt{x}$$ or,


 $$ (x-1)^2=x.$$


 This quadratic has solutions $x_{1,2}=\frac{3\pm\sqrt{5}}2$, and as we know $G>1$, we have $G=\frac{3+\sqrt{5}}2$ and ultimately


$$f(3)= 3g(3)=3G=\frac32(3+\sqrt{5}).$$

