# How do you write (5k^3)/k^-9 using only positive exponents?

$\textcolor{b l u e}{5 {k}^{12}}$
$\frac{5 {k}^{3}}{k} ^ \left(- 9\right)$
${k}^{- 9} = \frac{1}{k} ^ 9$
$\frac{5 {k}^{3}}{\frac{1}{k} ^ 9} = 5 {k}^{3} {k}^{9} = 5 {k}^{12}$