- If a is real number and \(\displaystyle\frac{m}{{n}} \text{is rational number in lowest terms with}\ \displaystyle{n}>{1},\ \text{then}\ \displaystyle{\sqrt[{{n}}]{{{a}^{m}}}}={\left({a}^{m}\right)}^{{\frac{1}{{n}}}}={a}^{{\frac{m}{{n}}}}\ \text{provided that}\ \displaystyle{\sqrt[{{n}}]{{a}}}\) is a real number.

- If a is real number and n is an integer with \(\displaystyle{n}\ge{2},\ \text{then}\ \displaystyle{a}^{{\frac{1}{{n}}}}={\sqrt[{{n}}]{{a}}}\ \text{provided that}\ \displaystyle{\sqrt[{{n}}]{{a}}}\) exists.

\(\displaystyle-{\left({a},{b}\right)}^{r}={a}^{r},{b}^{r}\)

- Power rule \(\displaystyle{\left({a}^{m}\right)}^{n}={a}^{{{m}{n}}}\)

Calculation:

Consider the expression \(\displaystyle\sqrt{{{81}{a}^{12}{b}^{20}}}\)

Write the expression \(\displaystyle\sqrt{{{81}{a}^{12}{b}^{20}}}\) using rational exponents.

And we know that \(\displaystyle{\sqrt[{{n}}]{{{a}^{m}}}}={\left({a}^{m}\right)}^{{\frac{1}{{n}}}}={a}^{{\frac{m}{{n}}}}\)

\(\displaystyle\sqrt{{{81}{a}^{12}{b}^{20}}}={\left({81}{a}^{12}{b}^{20}\right)}^{{\frac{1}{{2}}}}\)

From \(\displaystyle{\left({a},{b}\right)}^{r}={a}^{r},{b}^{r},{\left({81}{a}^{12}{b}^{20}\right)}^{{\frac{1}{{2}}}}={81}^{{\frac{1}{{2}}}}{\left({a}^{12}\right)}^{{\frac{1}{{2}}}},{\left({b}^{20}\right)}^{{\frac{1}{{2}}}}\)

\(\displaystyle{81}^{{\frac{1}{{2}}}}{\left({a}^{12}\right)}^{{\frac{1}{{2}}}},{\left({b}^{20}\right)}^{{\frac{1}{{2}}}}={\left({9}^{2}\right)}^{{\frac{1}{{2}}}}{\left({a}^{12}\right)}^{{\frac{1}{{2}}}},{\left({b}^{20}\right)}^{{\frac{1}{{2}}}}\)

And from power rule \(\displaystyle{\left({a}^{m}\right)}^{n}={a}^{{{m}{n}}}\ \text{we can write}\ \displaystyle{\left({9}^{2}\right)}^{{\frac{1}{{2}}}}{\left({a}^{12}\right)}^{{\frac{1}{{2}}}},{\left({b}^{20}\right)}^{{\frac{1}{{2}}}}\) as

\(\displaystyle{\left({9}^{2}\right)}^{{\frac{1}{{2}}}}{\left({a}^{12}\right)}^{{\frac{1}{{2}}}},{\left({b}^{20}\right)}^{{\frac{1}{{2}}}}={9},{a}^{6},{b}^{10}\)

Conclusion:

Therefore, \(\displaystyle\sqrt{{{81}{a}^{12}{b}^{20}}}={9}{a}^{6}{b}^{10}.\)