( a n ) m = a n ⋅ m a 1 m = a m {\displaystyle {(a^{n})}^{m}=a^{n\cdot m}\qquad a^{1 \over m}={\sqrt[{m}]{a}}}
( b 3 7 ) 28 9 {\displaystyle {\left(b^{3 \over 7}\right)}^{28 \over 9}} = b 3 7 ⋅ 28 9 = b 4 3 ( = b 4 3 ) {\displaystyle \quad =b^{{\frac {3}{7}}\cdot {\frac {28}{9}}}=b^{\frac {4}{3}}\left(={\sqrt[{3}]{b^{4}}}\right)}
( c − 5 ) 3 {\displaystyle \quad {(c^{-5})}^{3}} = a ( − 5 ) ⋅ 3 = c − 15 ( s − 5 ) − 4 = s ( − 5 ) ⋅ ( − 4 ) = s 20 {\displaystyle \quad =a^{(-5)\cdot 3}=c^{-15}\qquad {(s^{-5})}^{-4}=s^{(-5)\cdot (-4)}=s^{20}}
( a 7 ) 3 {\displaystyle \quad {(a^{7})}^{3}} = a 7 ⋅ 3 = a 21 ( x b ) 4 = x b ⋅ 4 = x 4 b ( x x ) 4 = x x ⋅ 4 = x 4 x {\displaystyle \quad =a^{7\cdot 3}=a^{21}\qquad {(x^{b})}^{4}=x^{b\ \cdot \ 4}=x^{4b}\qquad {(x^{x})}^{4}=x^{x\ \cdot \ 4}=x^{4x}}
( w − 5 35 ) − 14 {\displaystyle {\left({\sqrt[{35}]{w^{-5}}}\right)}^{-14}} = [ ( w − 5 ) 1 35 ] − 14 = w ( − 5 ) ⋅ 1 35 ⋅ ( − 14 ) = w 70 35 = w 2 {\displaystyle \quad ={\left[{(w^{-5})}^{1 \over 35}\right]}^{-14}=w^{(-5)\cdot {1 \over 35}\cdot {(-14)}}=w^{70 \over 35}=w^{2}}
y − 5 4 3 {\displaystyle {\sqrt[{4}]{y^{-5}}}^{\ 3}} = [ ( y − 5 ) 1 4 ] 3 = y ( − 5 ) ⋅ 1 4 ⋅ 3 = y − 15 4 ( = y 3 , 75 ) {\displaystyle \quad ={\left[{(y^{-5})}^{1 \over 4}\right]}^{3}=y^{(-5)\cdot {1 \over 4}\cdot 3}=y^{-{15 \over 4}}\left(=y^{3{,}75}\right)}
( ( b 3 5 ) 3 ⋅ b 6 5 ⋅ b − 2 ) 29 {\displaystyle \left(\left(b^{3 \over 5}\right)^{3}\cdot {\sqrt[{5}]{b^{6}}}\cdot b^{-2}\right)^{29}} = ( b 3 5 ⋅ 3 ⋅ b 6 5 ⋅ b − 2 ) 29 = ( b 9 5 + 6 5 − 2 ) 29 = … {\displaystyle \quad =\left(b^{{\frac {3}{5}}\cdot 3}\cdot {b^{6 \over 5}}\cdot b^{-2}\right)^{29}=\left(b^{{\frac {9}{5}}+{\frac {6}{5}}-2}\right)^{29}=\dots }
⋯ = ( b 15 5 3 − 2 ) 29 = ( b 1 ) 29 = b 29 {\displaystyle \dots =\left(b^{{\cancelto {3}{\ \ {\frac {15}{5}}\ \ }}-2}\right)^{29}=\left(b^{1}\right)^{29}=b^{29}}
( b 15 7 ) 28 3 b 3 b − 8 {\displaystyle {\dfrac {\sqrt[{3}]{{\Bigl (}b^{15 \over 7}{\Bigr )}^{28}}}{b^{3} \over b^{-8}}}} = ( b 15 7 ) 28 3 b 3 − ( − 8 ) = b 15 7 ⋅ 28 3 b 3 + 8 = b 20 b 11 = b 9 {\displaystyle \quad ={\frac {\left(b^{\frac {15}{7}}\right)^{28 \over 3}}{b^{3-(-8)}}}={\frac {b^{{\frac {15}{7}}\cdot {\frac {28}{3}}}}{b^{3+8}}}={\frac {b^{20}}{b^{11}}}=b^{9}}
![{\displaystyle {(a^{n})}^{m}=a^{n\cdot m}\qquad a^{1 \over m}={\sqrt[{m}]{a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9305f4603aeaa21979d5d599e5e2a47c3822eb93)
![{\displaystyle \quad =b^{{\frac {3}{7}}\cdot {\frac {28}{9}}}=b^{\frac {4}{3}}\left(={\sqrt[{3}]{b^{4}}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef0d3ae848ee5f9d24852cfd9f76a034d811b5a9)
![{\displaystyle {\left({\sqrt[{35}]{w^{-5}}}\right)}^{-14}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c06f56fbd1a262653feef54702175bb87eabb3ed)
![{\displaystyle \quad ={\left[{(w^{-5})}^{1 \over 35}\right]}^{-14}=w^{(-5)\cdot {1 \over 35}\cdot {(-14)}}=w^{70 \over 35}=w^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfa256dce560697a676bb2a7d71d2f0fa40a53d8)
![{\displaystyle {\sqrt[{4}]{y^{-5}}}^{\ 3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77a8e2b6c9abd7a9c941d91cff75fbea7c69228b)
![{\displaystyle \quad ={\left[{(y^{-5})}^{1 \over 4}\right]}^{3}=y^{(-5)\cdot {1 \over 4}\cdot 3}=y^{-{15 \over 4}}\left(=y^{3{,}75}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ce500492f6e74903dc278ee8e84d63e89f24467)
![{\displaystyle \left(\left(b^{3 \over 5}\right)^{3}\cdot {\sqrt[{5}]{b^{6}}}\cdot b^{-2}\right)^{29}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb2517ff02af39a1bd559126d32914e2770b6674)
![{\displaystyle {\dfrac {\sqrt[{3}]{{\Bigl (}b^{15 \over 7}{\Bigr )}^{28}}}{b^{3} \over b^{-8}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95d161d96512b9be5236acfbd39fe3cbacfba1b9)
