# How do you simplify m^5/m^2 and write it using only positive exponents?

May 18, 2017

${m}^{3}$

#### Explanation:

${m}^{5} / {m}^{2} = \frac{m \times m \times m \times m \times m}{m \times m}$

$\textcolor{w h i t e}{{m}^{5} / {m}^{2}} = \frac{\cancel{m} \times \cancel{m} \times m \times m \times m}{\cancel{m} \times \cancel{m}} \leftarrow \text{ cancelling}$

$\textcolor{w h i t e}{{m}^{5} / {m}^{2}} = {m}^{3}$

$\text{this result can be expressed generally as}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{a}^{m} / {a}^{n} = {a}^{\left(m - n\right)}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\Rightarrow {m}^{5} / {m}^{2} = {m}^{\left(5 - 2\right)} = {m}^{3}$

May 18, 2017

${m}^{5} / {m}^{2} = {m}^{5 - 2} = {m}^{3}$

#### Explanation:

Before we just blindly apply a rule, let's look at what we we have and what it actually means.

${m}^{5} / {m}^{2} = \frac{m \cdot m \cdot m \cdot m \cdot m}{m \cdot m}$

We know that anything divided by itself is $1$, (except for $0$)
so each time we have $\frac{m}{m}$ it can be regarded as $1$

$\textcolor{b l u e}{\frac{m}{m} = \frac{\cancel{m}}{\cancel{m}} = 1}$

${m}^{5} / {m}^{2} = \frac{\textcolor{b l u e}{{\cancel{m}}^{1}} \cdot \textcolor{red}{{\cancel{m}}^{1}} \cdot m \cdot m \cdot m}{\textcolor{b l u e}{\cancel{m}} \cdot \textcolor{red}{\cancel{m}}}$

Both the $m$ factors in the denominator have cancelled out, leaving the denominator as $1$, but there are three $m$ factors still on top.

$\therefore {m}^{5} / {m}^{2} = \frac{{\cancel{m}}^{1} \cdot {\cancel{m}}^{1} \cdot m \cdot m \cdot m}{\cancel{m} \cdot \cancel{m}} = {m}^{3}$

This leads us to the rule for dividing with indices:

$\textcolor{red}{\text{If you are dividing and the bases are the same}}$, $\textcolor{red}{\text{subtract the indices of like bases.}}$

${m}^{5} / {m}^{2} = {m}^{5 - 2} = {m}^{3}$

Here is another example for clarity.

(x^6y^2z^5)/(x^3y^7z^4) = (x^3z)/y^5" "(larr"there were more on top")/(larr"there were more at the bottom")