# Is there a rule for the $n$th root of a radical?

The square root of $$n^2$$ is just $$n$$. e.g.

$$\sqrt{16^2}$$ = 16.

Is it the case that if you have $$\sqrt[n]{x^m}$$ it's the equivalent of $$\sqrt[n-m]{x}$$?

Examples:

$$\sqrt[4]{16^2}$$ = 4

$$\sqrt[4-2]{16}$$ also = 4 # So the rule looks like it works here.

However, if I have:

$$\sqrt[5]{32^2}$$ = 4 # from my calculator

But:

$$\sqrt[5-2]{32}$$ = $$\sqrt[3]{32}$$ = 3.17, not 4.

If I see a expression of the form $$\sqrt[n]{x^m}$$ is there a rule of algebra that I can apply to simplify?

• It's division, not subtraction... $\sqrt[\frac52]{32}=32^{\frac25}=4$. – abiessu Mar 2 at 15:29
• Keep in mind that $\sqrt{16^2}$ is literally $\sqrt[2]{16^2},$ so if the rule were $\sqrt[n-m]{x}$ then $\sqrt[2]{16^2}$ would be $\sqrt[2-2]{16} = \sqrt[0]{16} = 1.$ – David K Mar 2 at 15:31

$$\sqrt[n]{x^m} = \sqrt[n/m]x = x^{m/n}$$
It is a coincidence that $$4-2=4/2$$ so you got the right result.
Turn those radical signs into exponents, then use the general rule that $$(x^a)^b = x^{ab}.$$ So $$\sqrt[5]{32^2} = (32^2)^{1/5} = 32^{2/5} = (2^5)^{2/5} = 2^2 = 4.$$ No need for a calculator.
In general $$\sqrt[n]{x}=x^{\tfrac1n}$$, so $$\sqrt[n]{x^m}=(x^m)^{\tfrac1n}=x^{\tfrac mn}=\sqrt[n/m]{x}.$$