The first step to do is simplify $\sqrt{9 \sqrt{27 \sqrt{128}}}$

$\sqrt{9 \sqrt{27 \sqrt{128}}}$

= $\sqrt{9 \sqrt{27 \times 8 \sqrt{2}}}$

= $\sqrt{9 \sqrt{216\sqrt{2}}}$

= $\sqrt{9 \times 6 \sqrt{6\sqrt{2}}}$

= $\sqrt{54 \sqrt{6\sqrt{2}}}$

= $3\sqrt{6 \sqrt{6\sqrt{2}}}$

By substituting $\sqrt{9 \sqrt{27 \sqrt{128}}} = 3\sqrt{6 \sqrt{6\sqrt{2}}}$

We will get,

$3\sqrt{6 \sqrt{6\sqrt{2\times 3\sqrt{6 \sqrt{6\sqrt{2 \ldots}}}}}}$

= $3 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt \ldots }}}}}}$

Let $x = \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt \ldots }}}}}}$

$x^2 =6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt \ldots }}}}}$

So, $6x = x^2$

$x=6$

$3x= 3 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt \ldots }}}}}}$

So,

$3 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \sqrt \ldots }}}}}} = \boxed {18}$

So the answer is 18

$x = \sqrt {9\sqrt {27\sqrt {128\sqrt{9\sqrt {\ldots}}}}}$

$x = \sqrt {9\sqrt {27\sqrt {128x}}}$

$\frac {x^{2}}{9} = \sqrt {27\sqrt {128x}}$

$\frac {x^{4}}{81\cdot27} = 8\sqrt {2x}$

$x^{7} = 3^{14}\cdot2^{7}$

$x = 3^{2}\cdot2 = \boxed {18}$ .

Let the solution be $x$ , since the term occurs recursively it could be rewritten as:

$\sqrt{9 \sqrt{27 \sqrt{128x}}} = x =>$

By raising the equation to power 8:

$9^4 * 27^2 * 128x = x^8 =>$

$3^{14} * 2^7 = x^7 =>$

$x = 3^2 * 2 = \boxed{18}.$

This is my first time posting a solution and using LaTeX, so expect my solution to be bad :/

Let: $\sqrt{9\sqrt{27\sqrt{128\sqrt{9\sqrt{27\sqrt{128\sqrt{...}}}}}}} = x$ Then $9\sqrt{27\sqrt{128\sqrt{9\sqrt{27\sqrt{128\sqrt{...}}}}}} = x^{2}$ $\sqrt{27\sqrt{128\sqrt{9\sqrt{27\sqrt{128\sqrt{...}}}}}} = \frac{x^{2}}{9}$ $27\sqrt{128\sqrt{9\sqrt{27\sqrt{128\sqrt{...}}}}} = \frac{x^{4}}{9^{2}}$ $\sqrt{128\sqrt{9\sqrt{27\sqrt{128\sqrt{...}}}}} = \frac{x^{4}}{9^{2} \times 27}$ $128\sqrt{9\sqrt{27\sqrt{128\sqrt{...}}}} = \frac{x^{8}}{9^{4} \times 27^{2}}$ $\sqrt{9\sqrt{27\sqrt{128\sqrt{...}}}} = \frac{x^{8}}{9^{4} \times 27^{2} \times 128}$ Deriving from the first equation gives us: $x = \frac{x^{8}}{9^{4} \times 27^{2} \times 128}$ $9^{4} \times 27^{2} \times 128 = x^{7}$ $3^{14} \times 2^{7} = x^{7}$ $3^{2} \times 2 = x$ $\boxed{18} = x$

Let $x = \sqrt{9 \sqrt{27 \sqrt{128 \sqrt{9 \sqrt{27 \sqrt{128 \sqrt{....}}}}}}}$

$x = \sqrt{9 \sqrt{27 \sqrt{128x}}}$

Squaring both sides,

$x^{2} = 9 \sqrt{27 \sqrt{128x}}$

Squaring both sides,

$x^{4} = 9^{2} . 27 \sqrt{128x}$

Squaring both sides,

$x^{8} = 9^{4} . 27^{2} . 128x$

Since $x = 0$ can't be the solution, we have

$x^{7} = 9^{4} . 27^{2} . 128$

$x^{7} = 3^{8} . 3^{6} . 2^{7}$

$x^{7} = 3^{16} . 2^{7}$

Taking seventh root on both sides,

$x = 3^{2} . 2 = 18$

$\sqrt{9\sqrt{27\sqrt{128\sqrt{9\sqrt{27\sqrt{128\sqrt{\ldots}}}}}}} \Rightarrow \sqrt{3^2\sqrt{3^3\sqrt{2^7\sqrt{3^2\sqrt{3^3\sqrt{2^7\sqrt{\ldots}}}}}}}$ is equal to $3\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{\ldots}}}}}}$

Now I want to demostrate that $\sqrt{x\sqrt{x\sqrt{x\sqrt{\ldots}}}} = x$ :

$\sqrt{x\sqrt{x\sqrt{x\sqrt{\ldots}}}} = \displaystyle \prod_{i=1}^{\infty} x^{2^{-i}}$

I know also that $\hspace{0.3cm} \log \displaystyle \prod_{}^{} = \displaystyle \sum_{}^{} \log \hspace{0.3cm}$ because of the properties of logarithms

so $\hspace{0.5cm} log_a \displaystyle \prod_{i=1}^{\infty} x^{2^{-i}} = \displaystyle \sum_{i=1}^{\infty} \log_a (x^{2^{-i}}) \hspace{0.5cm}$ for $a > 0$ & $a \neq 1$

$\Rightarrow \displaystyle \sum_{i=1}^{\infty} 2^{-i}\log_a (x) = \log_a (x)\displaystyle \sum_{i=1}^{\infty} 2^{-i} = \log_a(x)$ since $\hspace{0.5cm} \boxed{ \displaystyle \sum_{i=1}^{\infty} 2^{-i} = 1}$

so $\hspace{0.5cm} \ log_a \displaystyle \prod_{i=1}^{\infty} x^{2^{-i}} = \log_a(x)$

$\Rightarrow \displaystyle \prod_{i=1}^{\infty} x^{2^{-i}} = x$

so $\sqrt{x\sqrt{x\sqrt{x\sqrt{\ldots}}}} = x$

so $\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{\ldots}}}}}} = 6$

and finally $\hspace{0.5cm} 3\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{\ldots}}}}}} = 3 * 6 = 18$

Let's call $\sqrt {9\sqrt {27\sqrt {128\sqrt {9\sqrt {27\sqrt {128\sqrt {...}}}}}}}=P$ then

$\sqrt {9\sqrt {27\sqrt {128P}}}=P$

Continuing,

$9\sqrt {27\sqrt {128P}}=P^2$

$\sqrt {27\sqrt {128P}}=\frac {P^2}{9}$

$27\sqrt {128P}=\frac {P^4}{3^4}$

$\sqrt {128P}=\frac {P^4}{3^7}$

$128P=\frac {P^8}{3^{14}}$

$P=\frac {P^8}{2^7\times3^{14}}$

$P^7=2^7\times3^{14}$

$P=2\times3^2=\boxed {18}$

Let $y=\sqrt{9 \sqrt{27 \sqrt{128 \sqrt{9 \sqrt{27 \sqrt{128 \sqrt{...}}}}}}}$

Then, we can write

$y=\sqrt{9 \sqrt{27 \sqrt{128 \cdot y}}}$

$y^{2}=9 \sqrt{27 \sqrt{128 \cdot y}}$

$y^{4}=9^{2} \cdot 27 \sqrt{128 \cdot y}$

$y^{8}=9^{4} \cdot 27^{2} \cdot 128 \cdot y$

$y^{7}=9^{7} \cdot 2^{7}$

$y=2 \cdot 9=18$

For more information

Let the given surds = y, Then [9[27[128 y = y {{{ Let [ indicate square root sign}}}} Squaring both side we get, 9[27[128 y = y^2 or, 3^2[3^3[2^7 y = y^2 squaring again, we get, 3^4 3^3[2^7 y = y^4 squaring again, we get, 3^8 3^6 2^7 y = y^8 or, 3^8 3^6 2^7 = y^7 or, 3^14 2^7 = y^7 or, 9^7 2^7 = y^7 or,9*2=y i.e y=18

First, we notice that the sequence of square roots is recursive. We can thus write an equation: $x=\sqrt{9 \sqrt{27 \sqrt{128 x}}}$ This simplifies to: $x^{7}=612220032$ This equation has exactly 7 solutions, but only one of them is real. That solution is 18.

We use here the formula of sum of infinite Geometric Progression series when common ratio(r) < 1, Sum $(S)=\frac{a}{1-r}$ where $a$ is first term of GP and $r$ is the common ratio, i.e, ratio between 2 consecutive terms of the GP.

$\sqrt{9 \sqrt{27 \sqrt{128 \sqrt{9 \sqrt{27 \sqrt{128 \sqrt{....}}}}}}}$

$=9^{\frac{1}{2}}\times 27^{\frac{1}{4}}\times 128^{\frac{1}{8}}\times 9^{\frac{1}{16}}\times 27^{\frac{1}{32}}\times 128^{\frac{1}{64}}\times ........$

$=9^{(\frac{1}{2}+\frac{1}{16}+....)}\times 27^{(\frac{1}{4}+\frac{1}{32}+....)}\times 128^{(\frac{1}{8}+\frac{1}{64}+.....)}$

$=9^{\frac{\frac{1}{2}}{1-\frac{1}{8}}}\times 27^{\frac{\frac{1}{4}}{1-\frac{1}{8}}}\times 128^{\frac{\frac{1}{8}}{1-\frac{1}{8}}}$

$=9^{\frac{\frac{1}{2}}{\frac{7}{8}}}\times 27^{\frac{\frac{1}{4}}{\frac{7}{8}}}\times 128^{\frac{\frac{1}{8}}{\frac{7}{8}}}$

$=9^{\frac{4}{7}}\times 27^{\frac{2}{7}}\times 128^{\frac{1}{7}}$

$=(3^{2})^{\frac{4}{7}}\times (3^{3})^{\frac{2}{7}}\times (2^{7})^{\frac{1}{7}}$

$=(3)^{2\times \frac{4}{7}}\times (3)^{3\times \frac{2}{7}}\times (2)^{7\times \frac{1}{7}}$

$=(3)^{\frac{8}{7}}\times (3)^{\frac{6}{7}}\times (2)^{1}$

$=(3)^{\frac{8}{7}+\frac{6}{7}}\times 2$

$=(3)^{\frac{14}{7}}\times 2 = (3)^{2}\times 2 = 9\times 2 = \boxed{18}$

Let $x$ be $\sqrt{\color{#D61F06}{9}\sqrt{\color{#3D99F6}{27}\sqrt{\color{#20A900}{128}\sqrt{\color{#D61F06}{9}\sqrt{\color{#3D99F6}{27}\sqrt{\color{#20A900}{128}\sqrt{\cdots}}}}}}}$ . $\begin{array}{rcl} \sqrt{\color{#D61F06}{9}\sqrt{\color{#3D99F6}{27}\sqrt{\color{#20A900}{128}x}}}&=&x\\ \sqrt{\color{#D61F06}{3^2}\sqrt{\color{#3D99F6}{3^3}\sqrt{\color{#20A900}{2^7}x}}}&=&x\\ \color{#D61F06}{3^2}\sqrt{\color{#3D99F6}{3^3}\sqrt{\color{#20A900}{2^7}x}}&=&x^2\\ \sqrt{\color{#3D99F6}{3^3}\sqrt{\color{#20A900}{2^7}x}}&=&\frac{x^2}{\color{#D61F06}{3^2}}\\ \color{#3D99F6}{3^3}\sqrt{\color{#20A900}{2^7}x}&=&\frac{x^4}{\color{#D61F06}{3^4}}\\ \sqrt{\color{#20A900}{2^7}x}&=&\frac{x^4}{\color{#D61F06}{3^4}\cdot\color{#3D99F6}{3^3}}\\ \color{#20A900}{2^7}x&=&\frac{x^8}{\color{#D61F06}{3^8}\cdot\color{#3D99F6}{3^6}}\\ x&=&\frac{x^8}{\color{#D61F06}{3^8}\cdot\color{#3D99F6}{3^6}\cdot\color{#20A900}{2^7}}\\ x^7&=&\color{#69047E}{3^{14}}\cdot\color{#20A900}{2^7}\\ x&=&\color{#69047E}{3^2}\cdot\color{#20A900}2\\ x&=&\boxed{18}\\ \end{array}$

Let's call the expression $w$ and rewrite it as: $w=9^{1/2} \times 27^{1/4} \times 128^{1/8} \times 9^{1/16} \times 27^{1/32} \times 128^{1/64} \times \ldots$

If we raise it to the power of $8$ we get $w^{8}=9^{4} \times 27^{2} \times 128^{1} \times (9^{1/2} \times 27^{1/4} \times 128^{1/8} \times \ldots)$ which means $w^{8}=9^{4} \times 27^{2} \times 128^{1} \times w$ . Manipulating the exponents, we find the answer:

$w^{7}=3^{8} \times 3^{6} \times 2^{7} \rightarrow w^{7} = 3^{14} \times 2^{7} \rightarrow w = 3^{2} \times 2=\boxed{18}$

import math

k = 1
for i in range(200):
    k = math.sqrt(128*k)
    k = math.sqrt(27*k)
    k = math.sqrt(9*k)
print k

# Returns
# 18.0

Let the given expression shown be equal to $x$ . Then, from the pattern:

$x = \sqrt{9\sqrt{27\sqrt{128x}}}$

$\Longrightarrow x = 3\sqrt{3\sqrt{3 \times 8\sqrt{2x}}}$

$\Longrightarrow x = 3\sqrt{6\sqrt{6\sqrt{2x}}}$

Squaring both sides,

$\Longrightarrow x^{2} = 54\sqrt{6\sqrt{2x}}$

Squaring again:

$\Longrightarrow x^{4} = 54^{2} \cdot 6\sqrt{2x}$

$\Longrightarrow x^{8} = \left(2^{2} \cdot 3^{6}\right)^{2}\left(2 \cdot 3\right)^{2}(2x)$

$\Longrightarrow x^{7} = 2^{7} \cdot 3^{14}$

$\Longrightarrow x = 2 \cdot 3^{2} = \boxed{18}$

You can write the expression recursively:

$a(n+1)=\sqrt{9\sqrt{27\sqrt{128 a(n)}}}$

Which simplifies to

$a(n+1)=3^{7/4}2^{7/8}a(n)^{1/8}$

If the sequence has a limit, it will be a fixed point of the recursion. To find the fixed points just solve

$x=3^{7/4}2^{7/8} x^{1/8}$

There are two solutions, $x=0$ and $x=18$ .

$x=0$ is a trivial solution, thus the limit must be $18$ (assuming it exists).

try generalizing and making a common element having cube or square then just commionsense

i love pizza

we can do this by 2 methods.one of them is---

Let the value comes to be x.

now x=sqrt[9sqrt{27sqrt(128x)}]

squaring both the sides 3 times you will get---

x^7=3^14*2^7

x=3^2*2=18.

9=3 * 3,27=3 * 3 * 3,128=2 * 2 * 2 * 2 * 2 let calculate all 3 * 2 * 3=18

Guessed it right!

$x=\sqrt{9\sqrt{27\sqrt{128\sqrt{9\sqrt{27\sqrt{128...}}}}}}$

$x^2=9\sqrt{27\sqrt{128\sqrt{9\sqrt{27\sqrt{128...}}}}}$

$x^4=81*27\sqrt{128\sqrt{9\sqrt{27\sqrt{128...}}}}$

$x^8=81^2*27^2*128\sqrt{9\sqrt{27\sqrt{128...}}}$

$x^8=81^2*27^2*128*x$

$x=18$

let sqrt(9sqrt(27sqrt(128)))=x

then sqrt(9sqrt(27sqrt(128x)))=x

now on squaring both sides for three times we get

(81 27)^2 128=x^7

on solving we get x=18

let $\sqrt{9\sqrt{27\sqrt{128\sqrt{9\sqrt{27\sqrt{128\sqrt{...}}}}}}}=x$ and then we get $(9^2 27)^2 128 \sqrt{9\sqrt{27\sqrt{128\sqrt{...}}}}=x^8$ $(9^2 27)^2 128 x=x^8$ $(3^4 3^3)^2 128 x=x^8$ $3^{14} 2^7 =x^7$ $3^2 2 =x$ $x=18$

\sqrt{9sqrt{27sqrt{128sqrt x}}} = x. then, x^4 = [\sqrt{9sqrt{27sqrt{128sqrt x}}}]^4 x^4 = 81 27\sqrt{128} x^4 2 = [ 81 27\sqrt{128}]^2 x^8 = 6561 729 128 x = [6561 729*128]^8= 18 {if you don't understand it this way , just keep in mind that keep on squaring both the sides of the equation until all the numbers are out of the nested surd and then find the nth root of the final product , where 'n' is the exponent of the answer 'x'.]

First, we can observe that the first three nested surds are equal to

$\sqrt{9 \, \sqrt{27 \, \sqrt{128}}} = \sqrt[8]{18^{7}}$

So, our entire nested expression, let´s call it S , can be wrote as

$S = \sqrt[8]{18^{7} \, \sqrt[8]{18^{7} \, \sqrt[8]{18^{7} \, \sqrt[8]{...}}}}$

Thus

$S = 18^{\frac{7}{8}} \cdot 18^{\frac{7}{8^{2}}} \cdot 18^{\frac{7}{8^{3}}} \cdot 18^{\frac{7}{8^{4}}} \cdot ...$

$S = \left( 18^{7/8} \right)^{1 + \left( \frac{1}{8} \right) +\left( \frac{1}{8} \right)^{2} +\left( \frac{1}{8} \right)^{3} + {...}}$

$S = \left( 18^{7/8} \right)^{\frac{8}{7}}$

$S = 18$

take the given nested surds equal to y , then take power 8 on both sides we get y^8=(18)^7(y) or y^7=(18)^7 this gives y=18

what we can see is a pattern after every the under root things if we take it all equal to Y ,then we can get Y=3(SQ ROOT OF 6 AND INSIDE SQ ROOT OF 6 LIKE THAT) AFTER THAT SQUARE BOTH SIDE U WILL GET

Y SQ=9 * 6 * Y / 3

THEN Y=18

Y=/9/27/128y Y2=9/27/128y Y2/9=/27/128y (Y2/9)2=27/128y Y4/37=/128y Y8/314=128y Y7=27*97 Y=18

... É apenas uma questão de teoria , se levar em conta que o inicio até os pontinhos fica 3V^4 3^3.2^3V2.3V^4 3^3.2^3V2 ... ,podemos ver que os números que ficariam fora das raízes seriam aproximadamente 3.3.2 , visto que o 3 e o dois estão elevados a terceira e se repetem em dizima , então o numero aproximado seria 18 , bem não segui essa linha fiz de um jeito mais complicado , mas se esse sistema funcionar eu dou um ed avisando!

let y= y= and squaring three times so y^7=18^7