Question

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

Answer 1

`m^3`

Explanation:

`m^5/m^2=(mxxmxxmxxmxxm)/(mxxm)`

`color(white)(m^5/m^2)=(cancel(m)xxcancel(m)xxmxxmxxm)/(cancel(m)xxcancel(m))larr" cancelling"`

`color(white)(m^5/m^2)=m^3`

`"this result can be expressed generally as"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(a^m/a^n=a^((m-n)))color(white)(2/2)|)))`

`rArrm^5/m^2=m^((5-2))=m^3`

Answer 2

`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 = (m*m*m*m*m)/(m*m)`

We know that anything divided by itself is `1`, (except for `0`)
so each time we have `m/m` it can be regarded as `1`

`color(blue)(m/m = cancelm/cancelm=1)`

`m^5/m^2 = (color(blue)(cancelm^1)*color(red)(cancelm^1)*m*m*m)/(color(blue)(cancelm)*color(red)(cancelm))`

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

`:.m^5/m^2 = (cancelm^1*cancelm^1*m*m*m)/(cancelm*cancelm) =m^3`

This leads us to the rule for dividing with indices:

`color(red)("If you are dividing and the bases are the same")`, `color(red)("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")`