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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?

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    $\begingroup$ It's division, not subtraction... $\sqrt[\frac52]{32}=32^{\frac25}=4$. $\endgroup$ – abiessu Mar 2 at 15:29
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    $\begingroup$ 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 $\sqrt[2-2]{16} = \sqrt[0]{16} = 1.$ $\endgroup$ – David K Mar 2 at 15:31
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The rule would be

$$\sqrt[n]{x^m} = \sqrt[n/m]x = x^{m/n}$$

It is a coincidence that $4-2=4/2$ so you got the right result.

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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.

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In general $\sqrt[n]{x}=x^{\tfrac1n}$, so $$\sqrt[n]{x^m}=(x^m)^{\tfrac1n}=x^{\tfrac mn}=\sqrt[n/m]{x}.$$

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