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

2 Answers
May 18, 2017

Answer:

#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#

May 18, 2017

Answer:

#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")#