9.8 Rational exponents  (Page 2/7)

Solution

ⓐ
$\begin{array}{cccc}& & & {\left(\mathrm{-64}\right)}^{\frac{1}{3}}\hfill \\ \text{Rewrite as a cube root.}\hfill & & & \sqrt[3]{\mathrm{-64}}\hfill \\ \text{Rewrite}\mathrm{-64}\text{as a perfect cube.}\hfill & & & \sqrt[3]{{\left(\mathrm{-4}\right)}^3}\hfill \\ \text{Simplify.}\hfill & & & \mathrm{-4}\hfill \end{array}$

ⓑ
$\begin{array}{cccc}& & & \text{−}{64}^{\frac{1}{3}}\hfill \\ \text{The exponent applies only to the 64.}\hfill & & & \text{−}\left({64}^{\frac{1}{3}}\right)\hfill \\ \text{Rewrite as a cube root.}\hfill & & & \text{−}\sqrt[3]{64}\hfill \\ \text{Rewrite 64 as}{4}^3.\hfill & & & \text{−}\sqrt[3]{4^3}\hfill \\ \text{Simplify.}\hfill & & & \mathrm{-4}\hfill \end{array}$

ⓒ
$\begin{array}{cccc}& & & {\left(64\right)}^{-\frac{1}{3}}\hfill \\ \begin{array}{c}\text{Rewrite as a fraction with}\hfill \\ \text{a positive exponent, using}\hfill \\ \text{the property,}{a}^{\text{−}n}=\frac{1}{a^n}.\hfill \\ \text{Write as a cube root.}\hfill \end{array}\hfill & & & \frac{1}{\sqrt[3]{64}}\hfill \\ \text{Rewrite 64 as}{4}^3.\hfill & & & \frac{1}{\sqrt[3]{4^3}}\hfill \\ \text{Simplify.}\hfill & & & \frac{1}{4}\hfill \end{array}$

Solution

ⓐ
$\begin{array}{cccc}& & & {\left(\mathrm{-16}\right)}^{\frac{1}{4}}\hfill \\ \text{Rewrite as a fourth root.}\hfill & & & \sqrt[4]{\mathrm{-16}}\hfill \\ \text{There is no real number whose fourth power is}\mathrm{-16}.\hfill & & & \end{array}$

ⓑ
$\begin{array}{cccc}& & & \text{−}{16}^{\frac{1}{4}}\hfill \\ \begin{array}{c}\text{The exponent only applies to the 16.}\hfill \\ \text{Rewrite as a fourth root.}\hfill \end{array}\hfill & & & \text{−}\sqrt[4]{16}\hfill \\ \text{Rewrite}16\text{as}{2}^4.\hfill & & & \text{−}\sqrt[4]{2^4}\hfill \\ \text{Simplify.}\hfill & & & \mathrm{-2}\hfill \end{array}$

ⓒ
$\begin{array}{cccc}& & & {\left(16\right)}^{-\frac{1}{4}}\hfill \\ \text{Rewrite using the property}{a}^{\text{−}n}=\frac{1}{a^n}.\hfill & & & \frac{1}{{\left(16\right)}^{\frac{1}{4}}}\hfill \\ \text{Rewrite as a fourth root.}\hfill & & & \frac{1}{\sqrt[4]{16}}\hfill \\ \text{Rewrite}16\text{as}{2}^4.\hfill & & & \frac{1}{\sqrt[4]{2^4}}\hfill \\ \text{Simplify.}\hfill & & & \frac{1}{2}\hfill \end{array}$

Solution

We want to use $a^{\frac{m}{n}}=\sqrt[n]{a^m}$ to write each radical in the form $a^{\frac{m}{n}}$ .

1. ⓐ
   

   

2. ⓑ
   

   

3. ⓒ
   

   

Solution

We will rewrite each expression as a radical first using the property, $a^{\frac{m}{n}}={\left(\sqrt[n]{a}\right)}^m$ . This form lets us take the root first and so we keep the numbers in the radicand smaller than if we used the other form.

1. ⓐ
   $\begin{array}{cccc}& & & 9^{\frac{3}{2}}\hfill \\ \begin{array}{c}\text{The power of the radical is the numerator}\hfill \\ \text{of the exponent, 3. Since the denominator}\hfill \\ \text{of the exponent is 2, this is a square root.}\hfill \end{array}\hfill & & & {\left(\sqrt{9}\right)}^3\hfill \\ \\ \\ \text{Simplify.}\hfill & & & {\left(3\right)}^3\hfill \\ & & & 27\hfill \end{array}$
   
   
2. ⓑ
   $\begin{array}{cccc}& & & {125}^{\frac{2}{3}}\hfill \\ \begin{array}{c}\text{The power of the radical is the numerator}\hfill \\ \text{of the exponent, 2. The index of the radical}\hfill \\ \text{is the denominator of the exponent, 3.}\hfill \end{array}\hfill & & & {\left(\sqrt[3]{125}\right)}^2\hfill \\ \\ \\ \text{Simplify.}\hfill & & & {\left(5\right)}^2\hfill \\ & & & 25\hfill \end{array}$
   
   
3. ⓒ
   $\begin{array}{cccc}& & & {81}^{\frac{3}{4}}\hfill \\ \begin{array}{c}\text{The power of the radical is the numerator}\hfill \\ \text{of the exponent, 3. The index of the radical}\hfill \\ \text{is the denominator of the exponent, 4.}\hfill \end{array}\hfill & & & {\left(\sqrt[4]{81}\right)}^3\hfill \\ \\ \\ \text{Simplify.}\hfill & & & {\left(3\right)}^3\hfill \\ & & & 27\hfill \end{array}$

Solution

We will rewrite each expression first using $b^{\text{−}p}=\frac{1}{b^p}$ and then change to radical form.

ⓐ
$\begin{array}{cccc}& & & {16}^{-\frac{3}{2}}\hfill \\ \\ \\ \text{Rewrite using}{b}^{\text{−}p}=\frac{1}{b^p}.\hfill & & & \frac{1}{{16}^{\frac{3}{2}}}\hfill \\ \begin{array}{c}\text{Change to radical form. The power of the}\hfill \\ \text{radical is the numerator of the exponent, 3.}\hfill \\ \text{The index is the denominator of the}\hfill \\ \text{exponent, 2.}\hfill \end{array}\hfill & & & \frac{1}{{\left(\sqrt{16}\right)}^3}\hfill \\ \text{Simplify.}\hfill & & & \frac{1}{4^3}\hfill \\ \\ \\ & & & \frac{1}{64}\hfill \end{array}$

ⓑ
$\begin{array}{cccc}& & & {32}^{-\frac{2}{5}}\hfill \\ \\ \\ \text{Rewrite using}{b}^{\text{−}p}=\frac{1}{b^p}.\hfill & & & \frac{1}{{32}^{\frac{2}{5}}}\hfill \\ \\ \\ \text{Change to radical form.}\hfill & & & \frac{1}{{\left(\sqrt[5]{32}\right)}^2}\hfill \\ \\ \\ \text{Rewrite the radicand as a power.}\hfill & & & \frac{1}{{\left(\sqrt[5]{2^5}\right)}^2}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & \frac{1}{2^2}\hfill \\ & & & \frac{1}{4}\hfill \end{array}$

ⓒ
$\begin{array}{cccc}& & & 4^{-\frac{5}{2}}\hfill \\ \\ \\ \text{Rewrite using}{b}^{\text{−}p}=\frac{1}{b^p}.\hfill & & & \frac{1}{4^{\frac{5}{2}}}\hfill \\ \\ \\ \text{Change to radical form.}\hfill & & & \frac{1}{{\left(\sqrt{4}\right)}^5}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & \frac{1}{2^5}\hfill \\ & & & \frac{1}{32}\hfill \end{array}$