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

\({x^9 \over x^{-7}}=x^{9--7}=x^{16}\)

\(x^4⋅x^{-8}=x^{4+-8}=x^{-4}\)

\((a^6)^{-7}=a^{6⋅-7}=a^{-42}\)

\(a^3⋅{1 \over a^7}=a^3⋅a^{-7}=a^{3+-7}=a^{-4}\)

\({({1 \over x^6}) \over x^3}={x^{-6} \over x^3}=x^{-6-3}=x^{-9}\)

\({p^2 \over p^0}=p^{2-0}=p^2\)

\({7 \over x^4}\)

\({3a \over 8a^2}={3 \over 8}⋅{a \over a^2}={3 \over 8}⋅a^{1-2}={3 \over 8}a^{-1}\)

\({x^4 \over ({1 \over x^7})}={x^4 \over x^{-7}}=x^{4--7}=x^{11}\)

\({5p^5q^5 \over 2pq^{10}}={5 \over 2}⋅{p^5 \over p^1}⋅{q^5 \over q^{10}}={5 \over 2}⋅p^{5-1}⋅p^{5-10}=2\frac{1}{2}p^4q^{-5}\)

\(x^7⋅\sqrt[8]{x}=x^7⋅x^{\frac{1}{8}}=x^{7+\frac{1}{8}}=x^{7\frac{1}{8}}\)

\(a^7⋅\sqrt[5]{a^3}=a^7⋅a^{\frac{3}{5}}=a^{7+\frac{3}{5}}=a^{7\frac{3}{5}}\)

\({a^8 \over \sqrt[9]{a^5}}={a^8 \over a^{\frac{5}{9}}}=a^{8-\frac{5}{9}}=a^{7\frac{4}{9}}\)

\({1 \over x^7}⋅\sqrt[9]{x^4}=x^{-7}⋅x^{\frac{4}{9}}=x^{-7+\frac{4}{9}}=x^{-6\frac{5}{9}}\)

\({\sqrt[5]{p^4} \over \sqrt[7]{p^5}}={p^{\frac{4}{5}} \over p^{\frac{5}{7}}}=p^{\frac{4}{5}-\frac{5}{7}}=p^{\frac{3}{35}}\)

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

\(\sqrt[4]{x^8}=x^{\frac{8}{4}}=x^2\)

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

\({2q^9 \over 5p^2}\)

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

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

\(4a^{6\frac{5}{8}}=4⋅a^6⋅a^{\frac{5}{8}}=4a^6⋅\sqrt[8]{a^5}\)

\(\frac{2}{5}a^{-\frac{5}{6}}b^{\frac{1}{4}}=\frac{2}{5}⋅{1 \over a^{\frac{5}{6}}}⋅b^{\frac{1}{4}}={2⋅\sqrt[4]{b} \over 5⋅\sqrt[6]{a^5}}\)