# How do you simplify ((m^3p^5)/n^7)^6*((m^2n^0p^3)/(m^4n^2))^3 and write it using only positive exponents?

Mar 19, 2017

$\frac{{m}^{12} \textcolor{w h i t e}{.} {p}^{39}}{n} ^ \left(48\right)$

#### Explanation:

$\textcolor{b l u e}{\text{Preamble}}$

Consider the context of ${n}^{0}$
$\text{ }$suppose we had ${2}^{3} / {2}^{3} = 1$

$\text{ }$This can be written as ${2}^{3 - 3} = {2}^{0} = 1$

$\text{ }$ so ${n}^{0}$ disappears in the multiplication
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Consider the context of ${\left(\frac{{m}^{3} {p}^{5}}{n} ^ 7\right)}^{6}$

Suppose we had ${p}^{2}$ then this is $p \times p$

So whatever $p$ is it is multiplied by itself twice.

Set $p = {3}^{2}$ then we have ${3}^{2} \times {3}^{2} = 3 \times 3 \times 3 \times 3 = {3}^{4}$

So ${\left({3}^{2}\right)}^{2} = {3}^{2 \times 2} = {3}^{4}$
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$\textcolor{b l u e}{\text{Answering the question - applying the above principles}}$

Given:$\text{ } {\left(\frac{{m}^{3} {p}^{5}}{n} ^ 7\right)}^{6} \times {\left(\frac{{m}^{2} {n}^{0} {p}^{3}}{{m}^{4} {n}^{2}}\right)}^{3}$

$= \frac{{m}^{18} {p}^{30}}{n} ^ 42 \times \frac{{m}^{6} {p}^{9}}{{m}^{12} {n}^{6}}$

$= \frac{{m}^{18 + 6} \textcolor{w h i t e}{.} {p}^{30 + 9}}{{m}^{12} \textcolor{w h i t e}{.} {n}^{42 + 6}} \text{ }$ this is the same as: " "(m^12xxcancel(m^12))/(cancel(m^12)) xx(p^39)/(n^(48)

$= \frac{{m}^{12} \textcolor{w h i t e}{.} {p}^{39}}{n} ^ \left(48\right)$