\({1 \over a^2}=a^{-2}\)

\({a^4 \over a^{-2}}=a^{4--2}=a^6\)

\(x^4⋅x^{-7}=x^{4+-7}=x^{-3}\)

\((p^5)^{-8}=p^{5⋅-8}=p^{-40}\)

\(x^3⋅{1 \over x^5}=x^3⋅x^{-5}=x^{3+-5}=x^{-2}\)

\({({1 \over x^5}) \over x^3}={x^{-5} \over x^3}=x^{-5-3}=x^{-8}\)

\({p^0 \over p^7}=p^{0-7}=p^{-7}\)

\({5 \over x^6}\)

\({4a^5 \over 9a^8}={4 \over 9}⋅{a^5 \over a^8}={4 \over 9}⋅a^{5-8}={4 \over 9}a^{-3}\)

\({p^5 \over ({1 \over p^8})}={p^5 \over p^{-8}}=p^{5--8}=p^{13}\)

\({5a^4b^4 \over 4a^2b^7}={5 \over 4}⋅{a^4 \over a^2}⋅{b^4 \over b^7}={5 \over 4}⋅a^{4-2}⋅a^{4-7}=1\frac{1}{4}a^2b^{-3}\)

\(x^9⋅\sqrt[6]{x}=x^9⋅x^{\frac{1}{6}}=x^{9+\frac{1}{6}}=x^{9\frac{1}{6}}\)

\(p^8⋅\sqrt[7]{p^3}=p^8⋅p^{\frac{3}{7}}=p^{8+\frac{3}{7}}=p^{8\frac{3}{7}}\)

\({x^8 \over \sqrt[9]{x^7}}={x^8 \over x^{\frac{7}{9}}}=x^{8-\frac{7}{9}}=x^{7\frac{2}{9}}\)

\({1 \over a^2}⋅\sqrt[9]{a^8}=a^{-2}⋅a^{\frac{8}{9}}=a^{-2+\frac{8}{9}}=a^{-1\frac{1}{9}}\)

\({\sqrt[3]{a^2} \over \sqrt[5]{a^3}}={a^{\frac{2}{3}} \over a^{\frac{3}{5}}}=a^{\frac{2}{3}-\frac{3}{5}}=a^{\frac{1}{15}}\)

\(\sqrt[5]{{1 \over a^3}}=\sqrt[5]{a^{-3}}=a^{-\frac{3}{5}}\)

\(\sqrt[3]{x^{12}}=x^{\frac{12}{3}}=x^4\)

\({p^9 \over p^7⋅\sqrt[9]{p^7}}={p^9 \over p^7⋅p^{\frac{7}{9}}}={p^9 \over p^{7\frac{7}{9}}}=p^{9-7\frac{7}{9}}=p^{1\frac{2}{9}}\)

\({5y^6 \over 6x^7}\)

\((3a)^{-4}=3^{-4}⋅a^{-4}={1 \over 3^4}⋅{1 \over a^4}={1 \over 81a^4}\)

\(({1 \over 4}x)^{-2}=(4^{-1}⋅x)^{-2}=(4^{-1})^{-2}⋅x^{-2}=4^2⋅x^{-2}={16 \over x^2}\)

\(4x^{6\frac{4}{5}}=4⋅x^6⋅x^{\frac{4}{5}}=4x^6⋅\sqrt[5]{x^4}\)

\(\frac{7}{8}p^{-\frac{3}{5}}q^{\frac{2}{9}}=\frac{7}{8}⋅{1 \over p^{\frac{3}{5}}}⋅q^{\frac{2}{9}}={7⋅\sqrt[9]{q^2} \over 8⋅\sqrt[5]{p^3}}\)