How do you simplify $n^{6} \cdot {(n^{- 2})}^{5}$ $n^6 * (n^-2)^5$ ?

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How do you simplify $n^{6} \cdot {(n^{- 2})}^{5}$ $n^6 * (n^-2)^5$ ?

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Answer 1 · 2015-11-04T10:09:50

$n^{6} \cdot {(n^{- 2})}^{5} = \frac{1}{n^{4}}$ $n^6*(n^-2)^5=1/n^4$

Explanation:

$n^{6} \cdot {(n^{- 2})}^{5}$ $n^6*(n^-2)^5$

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

$n^{6} \cdot n^{- 2 \cdot 5} =$ $n^6*n^(-2*5)=$

$n^{6} \cdot n^{- 10}$ $n^6*n^-10$

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

$n^{6} \cdot n^{- 10} =$ $n^6*n^-10=$

$n^{6 \pm 10} =$ $n^(6+-10)=$

$n^{- 4}$ $n^-4$

Apply the exponent rule $a^{- m} = \frac{1}{a^{m}}$ $a^(-m)=1/a^m$ .

$n^{- 4} = \frac{1}{n^{4}}$ $n^-4=1/n^4$

Answer 2 · 2015-11-04T10:11:18

$\frac{1}{n^{4}}$ $1/n^4$

Explanation:

$n^{6} \cdot (n^{- 2 \cdot 5}) = n^{6} \cdot n^{- 10} = n^{6 + (- 10)} = n^{- 4} = \frac{1}{n^{4}}$ $n^6*(n^(-2*5))=n^6*n^(-10)=n^(6+(-10))=n^(-4)=1/n^4$

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How do you simplify $n^{6} \cdot {(n^{- 2})}^{5}$ $n^6 * (n^-2)^5$ ?

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