# How do you simplify (p^6*p^-4)^-2 leaving only positive exponents?

Apr 26, 2016

$\frac{1}{p} ^ 4$

#### Explanation:

Consider ${p}^{6} \times {p}^{- 4}$. This is the same as$\text{ } {p}^{6 - 4} = {p}^{2}$

So now we have ${\left({p}^{2}\right)}^{- 2}$

This is the same as $\text{ } \frac{1}{{\left({p}^{2}\right)}^{2}}$

Which is the same as $\text{ } \frac{1}{{p}^{2 \times 2}} = \frac{1}{p} ^ 4$

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$\textcolor{b l u e}{\text{Further explanation}}$

$\textcolor{b r o w n}{\text{Consider } {p}^{6} \times {p}^{- 4}}$

This is the same as $\frac{p \times p \times p \times p \times p \times p}{p \times p \times p \times p}$

Which is the same as $p \times p \times \frac{\cancel{p \times p \times p \times p}}{\cancel{p \times p \times p \times p}} \text{ "=" } {p}^{2} \times 1 = {p}^{2} \to {p}^{6 - 4}$

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color(brown)("Consider "(p^2)^2

This is ${\left(p \times p\right)}^{2} \to p \times p \times p \times p = {p}^{4} \to {p}^{2 \times 2}$

$\textcolor{b r o w n}{\text{But we need } {\left({p}^{2}\right)}^{- 2}}$

So $\textcolor{b l u e}{{\left({p}^{2}\right)}^{- 2} \to {\left[{\left({p}^{2}\right)}^{2}\right]}^{- 1} = {\left[{p}^{4}\right]}^{- 1} = \frac{1}{{p}^{4}}}$

Jun 17, 2016

$\frac{1}{p} ^ 4$

#### Explanation:

There is often more than one law of indices which can be used.
There is no correct order to apply the laws of indices, so there are often different methods of getting to the same answer.
Answers should not be left with negative of zero indices.

Instead of simplifying inside the bracket first, this can also be done by using the "power to a power" law, where you multiply the indices. This will remove the bracket.

${\left({p}^{6} \times {p}^{-} 4\right)}^{-} 2 = {p}^{-} 12 \times {p}^{8}$

There are now 2 choices:

Either - simplify by adding the indices and then attend to the negative index

${p}^{-} 12 \times {p}^{8} \text{ "= " "p^-4 " "= " } \frac{1}{p} ^ 4$

Or attend to the negative index first and then subtract the indices.

${p}^{-} 12 \times {p}^{8} \text{ "=" " 1/p^12 xx p^8" " = " } \frac{1}{p} ^ 4$