What is value of $\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\ldots}}}}=$?

Question

My question: how to find nested radical having $n$th roots $$\large\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\ldots}}}}=?$$ My try: $$\large\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\ldots}}}}=y$$ $$\large\sqrt[n]{xy}=y$$ $$\large xy=y^n$$$$ y^n-xy=0$$ Above $\mathrm{nth}$ degree equation should $n$ number of roots but my teacher says it has a unique value. I am totally confused how to get that value .

What will be the answer? I am looking for some trick or elegant way to solve such nested radical. Thanks

Answer 1

Here is a trick: $$\large \sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\ldots}}}=x^{\frac{1}{n}}\cdot x^{\frac{1}{n^2}}\cdot x^{\frac{1}{n^3}}\cdot\ldots=x^{\left(\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\ldots\right)}=x^{\frac{\frac1n}{1-\frac1n}}=x^{\frac{1}{n-1}} $$

Answer 2

Just do this in the last line. $$xy = y^n\iff \frac{y^n}{y} = y^{n-1} =x \iff \color{blue}{y = x^\frac{1}{n-1}}$$

Answer 3

Here is my answer:

I took $y$ common, $$y^n-xy=0$$ $$y(y^{n-1}-x)=0$$ $$y=0, \ \ y^{n-1}-x=0$$ $$y=0, \ \ y=x^{\dfrac1{n-1}}$$ But $y\ne 0$, I get the answer

$$ y=x^{\dfrac1{n-1}}$$