Answer
$\frac{a^{5}}{81}$
Work Step by Step
We are given the expression $\frac{9^{-5}a^{4}}{9^{-3}a^{-1}}$.
To simplify, we can use the quotient rule, which holds that $\frac{a^{m}}{a^{n}}=a^{m-n}$ (where a is a nonzero real number, and m and n are integers).
$(9^{-5-(-3)})\times(a^{4-(-1)})=9^{-2}\times a^{5}$
To simplify this into only positive exponents, we know that $a^{-n}=\frac{1}{a^{n}}$ (where a is a nonzero real number and n is a positive integer).
$\frac{a^{5}}{9^{2}}=\frac{a^{5}}{9\times9}=\frac{a^{5}}{81}$