What's the rule for it again?
a^-1/n = __________
a^-m/n= __________
a^-1/n = __________
a^-m/n= __________
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a^-1/n = 1/(nth root of a)
a^-m/n= 1/(nth root of a^m)
i.e. 3^-2/3=1/(cube root of 9, as 3^2=9)
a^-m/n= 1/(nth root of a^m)
i.e. 3^-2/3=1/(cube root of 9, as 3^2=9)
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To make negative indices positive, bring it to the bottom of a fraction. And to deal with the fraction, another way of writing it (and I think this is what you want me to answer) is to write it as a root.
So a^ (-1/n) = 1/ a^(1/n) = 1 / (nth root (a)) e.g. if n = 2, formula equals 1/ sqrt(a)
and a ^ (-m/n) = 1 / [ n root( a ^ m) ]
So a^ (-1/n) = 1/ a^(1/n) = 1 / (nth root (a)) e.g. if n = 2, formula equals 1/ sqrt(a)
and a ^ (-m/n) = 1 / [ n root( a ^ m) ]
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1).
a^-1/n
= 1/nth root of a
2).
a^-m/n
= 1/nth root of a^m
a^-1/n
= 1/nth root of a
2).
a^-m/n
= 1/nth root of a^m
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a^-1/n = 1/a^(1/n) = 1/(nth root of a)
a^-m/n= 1/a^(m/n) = 1/(nth root of a^m)
a^-m/n= 1/a^(m/n) = 1/(nth root of a^m)