Solving $\sqrt[3]{{x\sqrt {x\sqrt[3]{{x \ldots \sqrt x }}} }} = 2$ with 100 nested radicals

Question

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.

Answer 1

Well, $\sqrt[3]{x\sqrt{x}} = (x^{\frac 12}x)^{\frac 13}= (x^{1+\frac 12})^{\frac 13} = x^{\frac { 1+ \frac 12}3}=x^{\frac 13 + \frac 1{2\cdot 3}}$

And $\sqrt[3]{x\sqrt{x\sqrt[3]{x\sqrt{x}}}}= (x(x(x^{\frac 13 + \frac 1{2\cdot 3}}))^{\frac 12})^{\frac 13}=$

$(x(x^{1+\frac 13+ \frac 1{2\cdot 3}})^{\frac 12})^\frac 13=$

$(x(x^{\frac 12 + \frac 1{2\cdot 3}+\frac 1{2^2\cdot 3}})^{\frac 13}=$

$(x^{1+\frac 12 + \frac 1{2\cdot 3}+\frac 1{2^2\cdot 3}})^{\frac 13}=$

$x^{\frac 13+\frac 1{2\cdot 3} + \frac 1{2\cdot 3^2} + \frac 1{2^2\cdot 3^2}}$

We seem to be developing a pattern and we can almost see it and see why. We take the power so far, divide in half, add 1, and then divide in third.

If $\underbrace{\sqrt[3]{x\sqrt{x....}}}_{2k \text{ nested radicals}} = x^p$ then

$\sqrt[3]{x\sqrt{x{\underbrace{\sqrt[3]{x\sqrt{x....}}}_{2k \text{ nested radicals}}}}}=$

$x^{\frac {\frac{p+1}2+1}3}=x^{\frac 13+\frac 16 + \frac p6 }$

And that's the crux.

For $k=1$ we verified that for $2k$ nested radicals the expression is $x^{\frac 13 + \frac 16}$

And for $k=2$ we verified that for $2k$ nested radicals the expression is $x^{(\frac 12 + \frac 16) + \frac 16(\frac 12 + \frac 16)}$.

so if for $2k$ nested radicals we can assume the expression is $x^{p} = x^{(\frac 12 + \frac 16) + \frac 16(\frac 12 + \frac 16)+....... + \frac 1{6^{k-1}}(\frac 12 + \frac 16)}$.

then we have verifies for $2(k+1)$ the expression is $x^{(\frac 13 + \frac 16) + \frac 16p}= x^{(\frac 13 + \frac 16) + \frac 16(\frac 13 + \frac 16)+....... + \frac 1{6^{k}}(\frac 13 + \frac 16)}$.

ANd that is a proof by induction. We have proven by induction that $\underbrace{\sqrt[3]{x\sqrt{x....}}}_{2k \text{ nested radicals}} = x^{(\frac 12 + \frac 16) + \frac 16(\frac 13 + \frac 16)+....... + \frac 1{6^{k-1}}(\frac 13 + \frac 16)}$

And we can simplify that further:

$ x^{(\frac 13 + \frac 16) + \frac 16(\frac 13 + \frac 16)+....... + \frac 1{6^{k-1}}(\frac 12 + \frac 16)}=$

$x^{(\frac 13 + \frac 16)(1+....... + \frac 1{6^{k-1}})}$

ANd $\frac 13 + \frac 16 = \frac 12$ and $1+....... + \frac 1{6^{k-1}}= \frac {1-\frac 1{6^k}}{1-\frac 16}=(1-\frac 1{6^k})\frac 65$ and so $(\frac 13 + \frac 16)(1+....... + \frac 1{6^{k-1}})= \frac 12\cdot \frac 65(1-\frac 1{6^k})=\frac 35(1-\frac 1{6^k})$

So $x^{(\frac 13 + \frac 16)(1+....... + \frac 1{6^{k-1}})}= x^{\frac 35(1-\frac 1{6^k})}$

So we must solve $x^{\frac 35(1-\frac 1{6^{50}})}=2$.

So $x = 2^{\frac 1{\frac 35(1-\frac 1{6^{50}})}}=2^{\frac {5\cdot 6^{50}}{3(6^{50}-1)}}$

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

Here is an unannotated version of one possible solution:

$$\begin{array}{l} {E_1} = \sqrt[3]{{x\sqrt x }} = {x^{\frac{{{A_1}}}{6}}} = {x^{\frac{3}{6}}}, \\ {E_2} = \sqrt[3]{{x\sqrt {x\sqrt[3]{{x\sqrt x }}} }} = {x^{\frac{{{A_2}}}{{36}}}} = {x^{\frac{{21}}{{{6^2}}}}} = {x^{\frac{{6{A_1} + 3}}{{{6^2}}}}}, \\ {E_3} = \sqrt[3]{{x\sqrt {x\sqrt[3]{{x\sqrt {x\sqrt[3]{{x\sqrt x }}} }}} }} = {x^{\frac{{{A_3}}}{{216}}}} = {x^{\frac{{129}}{{{6^3}}}}} = {x^{\frac{{6{A_2} + 3}}{{{6^3}}}}}, \\ \ldots, \\ {E_{50}} = {x^{\frac{{{A_{50}}}}{{{6^{50}}}}}} = {x^{\frac{{6{A_{49}} + 3}}{{{6^50}}}}}, \\ B = 3, \\ q = 6, \\ d = 3, \\ {A_1} = B, \\ {A_{n + 1}} = q \times {A_n} + d, \\ {A_n} = {u_n} + C, \\ C = q \times C + d = d/(1 - q) = - \frac{3}{5}, \\ {u_1} = B - C, \\ {u_{n + 1}} = q \times {u_n}, \\ {u_n} = (B - C) \times {q^{n - 1}}, \\ {A_n} = (B - C) \times {q^{n - 1}} + C, \\ {A_{50}} = (3 + \frac{3}{5}) \times {6^{49}} - \frac{3}{5} = (6 \times \frac{3}{5}) \times {6^{49}} - \frac{3}{5} = \frac{3}{5} \times {6^{50}} - \frac{3}{5} = \frac{3}{5} \times ({6^{50}} - 1), \\ {x^{\frac{{\frac{3}{5} \times ({6^{50}} - 1)}}{{{6^{50}}}}}} = 2 \Rightarrow {x^{\frac{{3 \times ({6^{50}} - 1)}}{{5 \times {6^{50}}}}}} = 2 \Rightarrow x = {2^{\frac{{5 \times {6^{50}}}}{{3 \times ({6^{50}} - 1)}}}}. \end{array}$$