neato
alrighty, there are several ways to approach this
remmeber that
[tex]x^{-m}=\frac{1}{x^m}[/tex]
and
[tex](\frac{a}{b})^c=\frac{a^c}{b^c}[/tex]
and
[tex](ab)^c=(a^c)(b^c)[/tex]
and
[tex](a^b)^c=a^{bc}[/tex]
3 things we can do are
write a/b as ab⁻¹
or
distribute the -n to both a and b
or
just treat the whole thing as one fractoin
first way
[tex](\frac{a}{b})^{-n}=(ab^{-1})^{-n}=(a^{-n})(b^n)=(\frac{1}{a^n})(b^n)=\frac{b^n}{a^n}[/tex]
2nd way
[tex](\frac{a}{b})^{-n}=\frac{a^{-n}}{b^{-n}}=\frac{\frac{1}{a^n}}{\frac{1}{b^n}}=\frac{b^n}{a^n}[/tex]
3rd way
[tex](\frac{a}{b})^{-n}=\frac{1}{(\frac{a}{b})^n}=\frac{1}{\frac{a^n}{b^n}}=\frac{b^n}{a^n}[/tex]
in all cases, the answer is [tex]\frac{b^n}{a^n}[/tex]
