The given expression using rational exponents. Then simplify and convert back to radical notation. Assume that all variables represent positive real numbers. Given: The expression is displaystylesqrt{{{81}{a}^{12}{b}^{20}}}

The given expression using rational exponents. Then simplify and convert back to radical notation. Assume that all variables represent positive real numbers. Given: The expression is displaystylesqrt{{{81}{a}^{12}{b}^{20}}}

Question
The given expression using rational exponents. Then simplify and convert back to radical notation. Assume that all variables represent positive real numbers.
Given:
The expression is \(\displaystyle\sqrt{{{81}{a}^{12}{b}^{20}}}\)

Answers (1)

2021-02-06
Key points used:
- 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, \(\displaystyle\sqrt{{{81}{a}^{12}{b}^{20}}}={9}{a}^{6}{b}^{10}.\)
0

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