How do you simplify $4 n^{4} \cdot 2 n^{- 3}$ $4n^4*2n^-3$ and write it using only positive exponents?

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Answer 1 · 2018-03-04T09:49:23

$8 n^{1} = 8 n$ $8n^1=8n$

We have:

$4 n^{4} \cdot 2 n^{- 3}$ $4n^4*2n^-3$

$= 4 \cdot n^{4} \cdot 2 \cdot n^{- 3}$ $=4*n^4*2*n^-3$

$= 8 n^{4} n^{- 3}$ $=8n^4n^-3$

Using exponent rules, $a^{b} \cdot a^{c} = a^{b + c}$ $a^b*a^c=a^(b+c)$ .

So, we get

$= 8 n^{4 - 3}$ $=8n^(4-3)$

$= 8 n^{1}$ $=8n^1$

$= 8 n$ $=8n$

That's all!

How do you simplify 4n^4*2n^-34n4⋅2n−34n^4*2n^-3 and write it using only positive exponents?