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how would i solve rational exponent (y2/3)-9
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\(\bf \large{(a^n)^m\implies a^{n\cdot m}\qquad thus\\ \quad \\ \left(y^{\frac{2}{3}}\right)^{-9}\implies y^{\frac{2}{3}\cdot -9}}\)
ok i got that part
i would get y-18/3
\(\bf \large{ (a^n)^m\implies a^{n\cdot m}\qquad thus\\ \quad \\ \left(y^{\frac{2}{3}}\right)^{-9}\implies y^{\frac{2}{3}\cdot -9}\implies y^{-\frac{18}{3}}\implies y^{-6}}\)
do i simplify that and if so how would i write it
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\(\bf a^{-n} = \cfrac{1}{a^n}\qquad thus\\ \quad \\ y^{-6}\implies \cfrac{1}{y^6}\)
thats all i dont have to write a radical?
well, you didn't get a fraction in the exponent, so no radical, if it was ... say 1/2 or 5/6 then you'd end up with a root, but in this case you only have a "6"
ok
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