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

\({a^7 \over a^{-4}}=a^{7--4}=a^{11}\)

\(x^3⋅x^{-8}=x^{3+-8}=x^{-5}\)

\((x^4)^{-2}=x^{4⋅-2}=x^{-8}\)

\(a^5⋅{1 \over a^7}=a^5⋅a^{-7}=a^{5+-7}=a^{-2}\)

\({({1 \over x^6}) \over x^2}={x^{-6} \over x^2}=x^{-6-2}=x^{-8}\)

\({p^0 \over p^9}=p^{0-9}=p^{-9}\)

\({7 \over a^2}\)

\({5a^2 \over 8a^5}={5 \over 8}⋅{a^2 \over a^5}={5 \over 8}⋅a^{2-5}={5 \over 8}a^{-3}\)

\({x^2 \over ({1 \over x^6})}={x^2 \over x^{-6}}=x^{2--6}=x^8\)

\({7p^7q^2 \over 6p^5q^7}={7 \over 6}⋅{p^7 \over p^5}⋅{q^2 \over q^7}={7 \over 6}⋅p^{7-5}⋅p^{2-7}=1\frac{1}{6}p^2q^{-5}\)

\(a^2⋅\sqrt{a}=a^2⋅a^{\frac{1}{2}}=a^{2+\frac{1}{2}}=a^{2\frac{1}{2}}\)

\(a^3⋅\sqrt[9]{a^8}=a^3⋅a^{\frac{8}{9}}=a^{3+\frac{8}{9}}=a^{3\frac{8}{9}}\)

\({x^6 \over \sqrt[7]{x^3}}={x^6 \over x^{\frac{3}{7}}}=x^{6-\frac{3}{7}}=x^{5\frac{4}{7}}\)

\({1 \over x^2}⋅\sqrt[5]{x^4}=x^{-2}⋅x^{\frac{4}{5}}=x^{-2+\frac{4}{5}}=x^{-1\frac{1}{5}}\)

\({\sqrt[5]{a^4} \over \sqrt[9]{a^5}}={a^{\frac{4}{5}} \over a^{\frac{5}{9}}}=a^{\frac{4}{5}-\frac{5}{9}}=a^{\frac{11}{45}}\)

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

\(\sqrt[5]{a^{20}}=a^{\frac{20}{5}}=a^4\)

\({p^2 \over p^3⋅\sqrt[5]{p^4}}={p^2 \over p^3⋅p^{\frac{4}{5}}}={p^2 \over p^{3\frac{4}{5}}}=p^{2-3\frac{4}{5}}=p^{-1\frac{4}{5}}\)

\({2b^8 \over 9a^3}\)

\((3p)^{-5}=3^{-5}⋅p^{-5}={1 \over 3^5}⋅{1 \over p^5}={1 \over 243p^5}\)

\(({1 \over 3}x)^{-2}=(3^{-1}⋅x)^{-2}=(3^{-1})^{-2}⋅x^{-2}=3^2⋅x^{-2}={9 \over x^2}\)

\(8x^{2\frac{2}{7}}=8⋅x^2⋅x^{\frac{2}{7}}=8x^2⋅\sqrt[7]{x^2}\)

\(\frac{1}{8}p^{-\frac{7}{8}}q^{\frac{1}{4}}=\frac{1}{8}⋅{1 \over p^{\frac{7}{8}}}⋅q^{\frac{1}{4}}={1⋅\sqrt[4]{q} \over 8⋅\sqrt[8]{p^7}}\)