# How do you simplify (16^(5/9) * 5 ^ (7/9)) ^-3?

Mar 4, 2018

See a solution process below:

#### Explanation:

First, use this rule for exponents to remove the outer exponent:

${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

${\left({16}^{\textcolor{red}{\frac{5}{9}}} {5}^{\textcolor{red}{\frac{7}{9}}}\right)}^{\textcolor{b l u e}{- 3}} \implies$

${16}^{\textcolor{red}{\frac{5}{9}} \times \textcolor{b l u e}{- 3}} {5}^{\textcolor{red}{\frac{7}{9}} \times \textcolor{b l u e}{- 3}} \implies$

${16}^{\textcolor{red}{- \frac{5}{3}}} {5}^{\textcolor{red}{- \frac{7}{3}}}$

Next, we can use this rule of exponents to eliminate the negative exponents:

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

$\frac{1}{{16}^{\textcolor{red}{- - \frac{5}{3}}} {5}^{\textcolor{red}{- - \frac{7}{3}}}} \implies$

$\frac{1}{{16}^{\frac{5}{3}} {5}^{\frac{7}{3}}}$

Depending on how you are being requested to simplify the expression you can use this rule of exponents and radicals to write the expression in radical form:

${x}^{\frac{1}{\textcolor{red}{n}}} = \sqrt[\textcolor{red}{n}]{x}$

$\frac{1}{{16}^{5 \times \frac{1}{3}} {5}^{7 \times \frac{1}{3}}} \implies$

$\frac{1}{{\left({16}^{5}\right)}^{\frac{1}{3}} {\left({5}^{7}\right)}^{\frac{1}{3}}} \implies$

$\frac{1}{\sqrt[3]{{16}^{5}} \sqrt[3]{{5}^{7}}} \implies$

$\frac{1}{\sqrt[3]{{16}^{3} \cdot {16}^{2}} \sqrt[3]{{5}^{6} \cdot 5}} \implies$

$\frac{1}{\sqrt[3]{{16}^{3}} \sqrt[3]{{16}^{2}} \sqrt[3]{{5}^{6}} \sqrt[3]{5}} \implies$

$\frac{1}{16 \sqrt[3]{{16}^{2}} {5}^{2} \sqrt[3]{5}} \implies$

$\frac{1}{16 \times 25 \sqrt[3]{{16}^{2}} \sqrt[3]{5}} \implies$

$\frac{1}{400 \sqrt[3]{{16}^{2}} \sqrt[3]{5}} \implies$

$\frac{1}{400 \sqrt[3]{256} \sqrt[3]{5}} \implies$

$\frac{1}{400 \sqrt[3]{256 \times 5}} \implies$

$\frac{1}{400 \sqrt[3]{1280}}$