Question

Hello, how do I prove that root(m)(root(n)(x)) = root(mn)(x)?

I don't really understand it, thanks for helping!

Answer 1

\[\sqrt{m}\sqrt{nx} = \sqrt{mn} x\]

Is that the equation or does the "x" go inside the square root on the product side? I do not understand this question myself I am just making it easier to read for people who do.

Answer 2

I assume you mean the m'th root of x times the n'th root of x, yes?
\(\large\rm \sqrt[m]{x}\cdot\sqrt[n]{x}\)

Answer 3

oh oh i misread the question,
so like this?
\(\large\rm \sqrt[m]{\sqrt[n]{x}}\)

Answer 4

\[\sqrt[m]{\sqrt[n]{x}} = \sqrt[mn]{x}\] Yes

Answer 5

Well if you look at a^b^c, it's a well known rule of exponents that this will equal a^bc if you are taking the cth root of the bth root of a that is equal to a^(1/b)^(1/c) but if all we know is that a, b and c are integers we still have a^b^c therefore we will still have a^bc thus c*bth root of a.

Answer 6

\[\sqrt[m]{\sqrt[n]{x}} = \sqrt[n]{x}^{1/m} = x^{1/m \times 1/n} = x^{1/mn} = \sqrt[mn]{x}\]

Am I in the right direction?

Answer 7

Sure :) ya
I would use some brackets to be a little bit more clear though.
And work from the inside out like this,\[\large\rm \sqrt[m]{\color{orangered}{\sqrt[n]{x}}}\quad=\sqrt[m]{\color{orangered}{x^{1/n}}}\quad=\left(x^{1/n}\right)^{1/m}\]And by your exponent rule: \(\large\rm (x^a)^b=x^{ab}\)
We can say\[\large\rm =x^{\frac{1}{mn}}\]and rewrite it in root notation as you did,\[\large\rm =\sqrt[mn]{x}\]

Answer 8

Ah, thank you. Working inside out looks great and is beautifully done!