# How do you evaluate and simplify 9/9^(-4/5)?

Jul 9, 2017

See a solution process below:

#### Explanation:

First, use this rule of exponents to eliminate the negative exponent:

$\frac{1}{x} ^ \textcolor{red}{a} = {x}^{\textcolor{red}{- a}}$

$\frac{9}{9} ^ \textcolor{red}{- \frac{4}{5}} = 9 \cdot {9}^{\textcolor{red}{- - \frac{4}{5}}} = 9 \cdot {9}^{\frac{4}{5}}$

Next, use these rules to combine the 9's terms:

$a = {a}^{\textcolor{red}{1}}$ and ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$9 \cdot {9}^{\frac{4}{5}} \implies {9}^{\textcolor{red}{1}} \cdot {9}^{\textcolor{b l u e}{\frac{4}{5}}} \implies {9}^{\textcolor{red}{1} + \textcolor{b l u e}{\frac{4}{5}}} \implies {9}^{\textcolor{red}{\frac{5}{5}} + \textcolor{b l u e}{\frac{4}{5}}} \implies$

${9}^{\frac{9}{5}}$

Jul 9, 2017

:.color(blue)(=52.196 to the nearest 3 decimal places

#### Explanation:

$\frac{9}{\frac{1}{9} ^ \left(- \frac{4}{5}\right)}$

$\therefore \frac{1}{a} ^ - 2 = {a}^{2}$

$\therefore = \frac{9}{\frac{1}{9} ^ \left(\frac{4}{5}\right)}$

$\therefore = \frac{9}{1} \times {9}^{\frac{4}{5}} / 1$

$\therefore = {9}^{\frac{5}{5}} \times {9}^{\frac{4}{5}}$

$\therefore = {9}^{\frac{9}{5}}$

$\therefore = \sqrt[5]{{9}^{9}}$

$\therefore = \sqrt[5]{9 \cdot 9 \cdot 9 \cdot 9 \cdot 9 \cdot 9 \cdot 9 \cdot 9 \cdot 9}$

$\therefore \sqrt[5]{9} \cdot \sqrt[5]{9} \cdot \sqrt[5]{9} \cdot \sqrt[5]{9} \cdot \sqrt[5]{9} = 9$

$\therefore = 9 \sqrt[5]{9 \cdot 9 \cdot 9 \cdot 9}$

$\therefore = 9 \sqrt[5]{3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3}$

$\therefore \sqrt[5]{3} \cdot \sqrt[5]{3} \cdot \sqrt[5]{3} \cdot \sqrt[5]{3} \cdot \sqrt[5]{3} = 3$

$\therefore = 3 \cdot 9 \sqrt[5]{27}$

$\therefore = 27 \sqrt[5]{27}$

$\therefore = 52.19591521$

:.color(blue)(=52.196 to the nearest 3 decimal places