How do you simplify #( a^-2/a^5 )^-1#?

Question

1414 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-02T20:20:57

#a^7#

any number to the power of #-1# is the same as #1# divided by that number.

in other words, #a^-1 = 1/a^1#.

here we have #a^-2/a^5#

dividing #1# by #a^-2/a^5# gives #a^5/a^-2#.

it is also true that dividing one expression by another with the same base gives the base with a subtracted exponent.

in other words, #(a^m)/(a^n) = a^(m-n)#.

this means that #a^5/a^-2# is the same as #a^(5-(-2))#

#= a^(5+2)#

#= a^7#

hence, #((a^-2)/(a^5))^-1 = a^5/a^-2 = a^7#.