Solving $\sqrt[3]{{x\sqrt {x\sqrt[3]{{x \ldots \sqrt x }}} }} = 2$ with 100 nested radicals I have seen a book that offers to solve the following equation:
$$\underbrace {\sqrt[3]{{x\sqrt {x\sqrt[3]{{x \ldots \sqrt x }}} }}}_{{\text{100 radicals}}} = 2$$
The book also contains the answer:
$$x = {2^{\left( {\frac{{5 \times {6^{50}}}}{{3 \times ({6^{50}} - 1)}}} \right)}}$$
How did they get the answer for such equation? I tried to obtain the recurrence relation, but could not find the way to get the above answer.
EDIT
$${u_{100}} = 2,$$
$$\sqrt[3]{{x{u_{99}}}} = 2,$$
$$x{u_{99}} = {2^3},$$
$${u_{99}} = \sqrt {x{u_{98}}} ,$$
$$x\sqrt {x{u_{98}}}  = {2^3},$$
$${x^2}x{u_{98}} = {({2^3})^2},$$
$${x^3}{u_{98}} = {2^6},$$
$${u_{98}} = \sqrt[3]{{x{u_{97}}}},$$
$${x^3} \times \sqrt[3]{{x{u_{97}}}} = {2^6},$$
$${x^9}x{u_{97}} = {({2^6})^3},$$
$${x^{10}}{u_{97}} = {2^{18}},$$
$${u_{97}} = \sqrt {x{u_{96}}} ,$$
$${x^{10}}\sqrt {x{u_{96}}}  = {2^{18}},$$
$${x^{20}}x{u_{96}} = {({2^{18}})^2},$$
$${x^{21}}{u_{96}} = {2^{36}},$$
$${u_{96}} = \sqrt[3]{{x{u_{95}}}},$$
$${x^{21}} \times \sqrt[3]{{x{u_{95}}}} = {2^{36}},$$
$${x^{63}}x{u_{95}} = {2^{108}}$$
$${x^{64}}{u_{95}} = {2^{108}},$$
$${u_{95}} = \sqrt {x{u_{94}}} ,$$
$${x^{64}}\sqrt {x{u_{94}}}  = {2^{108}},$$
$${x^{128}}x{u_{94}} = {2^{216}},$$
$${x^{129}}{u_{94}} = {2^{216}},$$
$$ \ldots $$
but I still have no idea how to find a generalized formula which allows to obtain the answer.
 A: Essentially what is written there is just a big product of $x^{a_n}$ where $a_n$ changes from term to term. The terms $a_n$ follow the following sequence:
$$\frac{1}{3}, \frac{1}{3\cdot 2}, \frac{1}{3^2\cdot 2}, \frac{1}{3^2\cdot 2^2},\ldots, \frac{1}{3^{50}\cdot 2^{50}}$$
The equation you have now is
$$\prod_{n=1}^{100} x^{a_n} = 2$$
Which can be simplified to
$$x^{\left(\sum_{n=1}^{100} {a_n}\right)}= 2$$
The way the sequence $a_n$ is created allows you to say that $a_{2m-1}=2\cdot a_{2m}$, which gives
$$x^{3\left(\sum_{n=1}^{50} \frac{1}{6^n}\right)}= 2$$
This will now lead you directly to the solution
A: Let $u_n$ be the expression with $n$ radicals.  Then $u_{100} = 2$.  Then $u^3_{100} = xu_{99} = 2^3$.  Next $(xu_{99})^2 = x^3xu_{98} = 2^9$.  Keep going and try to find a pattern.
A: There is a simple trick:
$$\sqrt[\color{blue} A]{x^{\color{red}B}\sqrt[\color{blue}C]{x^{\color{red}D}\sqrt[\color{red}E]{x^{\color{blue} F}}  }} = x^{\left(((A\times B+C)\times D+E )\times F\right)/ACE}$$
This is  $ A~~\color{blue}{ times }~~ B~~ \color{blue}{ plus } ~~C~~  , \color{blue}{ times}~~ D ~~ \color{blue}{ plus } E, ~~\color{blue}{ times} ~~ F $
Then, you can find a rule for the $n-th$ term and figure out what will be your result
A: Here is an unannotated version of one possible solution:
$$\begin{array}{l}
{E_1} = \sqrt[3]{{x\sqrt x }} = {x^{\frac{{{A_1}}}{6}}} = {x^{\frac{3}{6}}}, \\
{E_2} = \sqrt[3]{{x\sqrt {x\sqrt[3]{{x\sqrt x }}} }} = {x^{\frac{{{A_2}}}{{36}}}} = {x^{\frac{{21}}{{{6^2}}}}} = {x^{\frac{{6{A_1} + 3}}{{{6^2}}}}}, \\
{E_3} = \sqrt[3]{{x\sqrt {x\sqrt[3]{{x\sqrt {x\sqrt[3]{{x\sqrt x }}} }}} }} = {x^{\frac{{{A_3}}}{{216}}}} = {x^{\frac{{129}}{{{6^3}}}}} = {x^{\frac{{6{A_2} + 3}}{{{6^3}}}}}, \\
\ldots, \\
{E_{50}} = {x^{\frac{{{A_{50}}}}{{{6^{50}}}}}} = {x^{\frac{{6{A_{49}} + 3}}{{{6^50}}}}}, \\
B = 3, \\
q = 6, \\
d = 3, \\
{A_1} = B, \\
{A_{n + 1}} = q \times {A_n} + d, \\
{A_n} = {u_n} + C, \\
C = q \times C + d = d/(1 - q) = - \frac{3}{5}, \\
{u_1} = B - C, \\
{u_{n + 1}} = q \times {u_n}, \\
{u_n} = (B - C) \times {q^{n - 1}}, \\
{A_n} = (B - C) \times {q^{n - 1}} + C, \\ 
{A_{50}} = (3 + \frac{3}{5}) \times {6^{49}} - \frac{3}{5} = (6 \times \frac{3}{5}) \times {6^{49}} - \frac{3}{5} = \frac{3}{5} \times {6^{50}} - \frac{3}{5} = \frac{3}{5} \times ({6^{50}} - 1), \\
{x^{\frac{{\frac{3}{5} \times ({6^{50}} - 1)}}{{{6^{50}}}}}} = 2 \Rightarrow {x^{\frac{{3 \times ({6^{50}} - 1)}}{{5 \times {6^{50}}}}}} = 2 \Rightarrow x = {2^{\frac{{5 \times {6^{50}}}}{{3 \times ({6^{50}} - 1)}}}}. 
\end{array}$$
