# How do you write 16^-9/16^-8 using only positive exponents?

Sep 5, 2016

$\frac{1}{16}$

#### Explanation:

We have: $\frac{{16}^{- 9}}{{16}^{- 8}}$

Using the laws of exponents:

$= {16}^{- 9 - \left(- 8\right)}$

$= {16}^{- 9 + 8}$

$= {16}^{- 1}$

$= \frac{1}{{16}^{1}}$

$= \frac{1}{16}$

Sep 5, 2016

$\frac{1}{16}$

#### Explanation:

Recall two of the laws of indices: Neither is more important - you can choose which to use.

${x}^{m} / {x}^{n} = {x}^{m - n} \mathmr{if} m > n \text{ but } {x}^{m} / {x}^{n} = \frac{1}{x} ^ \left(n - m\right) \mathmr{if} n > m$

${x}^{-} m = \frac{1}{x} ^ m$
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$\frac{\textcolor{red}{{16}^{-} 9}}{\textcolor{b l u e}{{16}^{-} 8}} \text{ } \leftarrow$ get rid of the negative indices

=$\frac{\textcolor{b l u e}{{16}^{8}}}{\textcolor{red}{{16}^{9}}}$

=$\frac{1}{16}$

OR:

${16}^{-} \frac{9}{16} ^ - 8 \text{ } \leftarrow$ subtract the indices:

=$\frac{1}{16} ^ \left(- 8 - \left(- 9\right)\right) = \frac{1}{16} ^ \left(- 8 + 9\right)$

=$\frac{1}{16}$