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

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