# How do you write 1/8 as a negative exponent?

$\frac{1}{8} = {8}^{- 1} = {2}^{- 3}$
Since from the laws of exponents we have that ${x}^{- n} = \frac{1}{x} ^ n$, it implies that we may write $\frac{1}{8} = {8}^{- 1}$.
$\frac{1}{8} = \frac{1}{2} ^ 3 = {2}^{- 3}$.