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

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

\(a^4⋅a^{-6}=a^{4+-6}=a^{-2}\)

\((a^4)^{-8}=a^{4⋅-8}=a^{-32}\)

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

\({({1 \over a^9}) \over a^7}={a^{-9} \over a^7}=a^{-9-7}=a^{-16}\)

\({p^3 \over p^0}=p^{3-0}=p^3\)

\({9 \over a^6}\)

\({3a^5 \over 5a^6}={3 \over 5}⋅{a^5 \over a^6}={3 \over 5}⋅a^{5-6}={3 \over 5}a^{-1}\)

\({x^4 \over ({1 \over x^8})}={x^4 \over x^{-8}}=x^{4--8}=x^{12}\)

\({8x^7y^3 \over 5x^3y^6}={8 \over 5}⋅{x^7 \over x^3}⋅{y^3 \over y^6}={8 \over 5}⋅x^{7-3}⋅x^{3-6}=1\frac{3}{5}x^4y^{-3}\)

\(p^5⋅\sqrt{p}=p^5⋅p^{\frac{1}{2}}=p^{5+\frac{1}{2}}=p^{5\frac{1}{2}}\)

\(a^8⋅\sqrt[9]{a^7}=a^8⋅a^{\frac{7}{9}}=a^{8+\frac{7}{9}}=a^{8\frac{7}{9}}\)

\({a^6 \over \sqrt[9]{a^2}}={a^6 \over a^{\frac{2}{9}}}=a^{6-\frac{2}{9}}=a^{5\frac{7}{9}}\)

\({1 \over x^3}⋅\sqrt[3]{x^2}=x^{-3}⋅x^{\frac{2}{3}}=x^{-3+\frac{2}{3}}=x^{-2\frac{1}{3}}\)

\({\sqrt[9]{a^8} \over \sqrt[5]{a^4}}={a^{\frac{8}{9}} \over a^{\frac{4}{5}}}=a^{\frac{8}{9}-\frac{4}{5}}=a^{\frac{4}{45}}\)

\(\sqrt[8]{{1 \over x^7}}=\sqrt[8]{x^{-7}}=x^{-\frac{7}{8}}\)

\(\sqrt[5]{p^{10}}=p^{\frac{10}{5}}=p^2\)

\({x^6 \over x^8⋅\sqrt[9]{x^2}}={x^6 \over x^8⋅x^{\frac{2}{9}}}={x^6 \over x^{8\frac{2}{9}}}=x^{6-8\frac{2}{9}}=x^{-2\frac{2}{9}}\)

\({8y^9 \over 9x^8}\)

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

\(({1 \over 2}a)^{-5}=(2^{-1}⋅a)^{-5}=(2^{-1})^{-5}⋅a^{-5}=2^5⋅a^{-5}={32 \over a^5}\)

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

\(\frac{1}{2}a^{-\frac{1}{9}}b^{\frac{1}{5}}=\frac{1}{2}⋅{1 \over a^{\frac{1}{9}}}⋅b^{\frac{1}{5}}={1⋅\sqrt[5]{b} \over 2⋅\sqrt[9]{a}}\)