How do you simplify $n^6 * (n^-2)^5$ ?

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Answer 1 · 2015-11-04T10:09:50

$n^6*(n^-2)^5=1/n^4$

$n^6*(n^-2)^5$

Simplify $(n^-2)^5$ by applying the exponent rule $(a^m)^n=a^(m*n)$ .

$n^6*n^(-2*5)=$

$n^6*n^-10$

Simplify by applying the exponent rule $a^m*a^n=a^(m+n)$ .

$n^6*n^-10=$

$n^(6+-10)=$

$n^-4$

Apply the exponent rule $a^(-m)=1/a^m$ .

$n^-4=1/n^4$

Answer 2 · 2015-11-04T10:11:18

$1/n^4$

$n^6*(n^(-2*5))=n^6*n^(-10)=n^(6+(-10))=n^(-4)=1/n^4$

How do you simplify n^6 * (n^-2)^5n^6 * (n^-2)^5?