How do you simplify #a^-2times a^3timesa^4#?

1 Answer
Mar 21, 2016

Answer:

#a^-2 xx a^3 xx a^4 = a^5#

Explanation:

If #n# is a positive integer then:

#a^n = overbrace(a xx a xx .. xx a)^"n times"#

So if #m# and #n# are both positive integers:

#a^m * a^n = overbrace(a xx a xx .. xx a)^"m times" xx overbrace(a xx a xx .. xx a)^"n times"#

#= overbrace(a xx a xx .. xx a)^"m + n times" = a^(m+n)#

If #a != 0# then we can also define:

#a^(-n) = 1/underbrace(a xx a xx .. xx a)_"n times"#

In fact, if #a != 0# then #a^m * a^n = a^(m+n)# regardless of whether #m# and #n# are positive, negative or zero.

#color(white)()#
In our example,

#a^-2 xx a^3 xx a^4 = a^(-2+3+4) = a^5#

Or if you prefer to see it a little slower:

#a^-2 xx a^3 xx a^4#

#=1/(a xx a) xx (a xx a xx a) xx (a xx a xx a xx a)#

#=(color(red)(cancel(color(black)(a xx a))) xx a xx a xx a xx a xx a)/color(red)(cancel(color(black)((a xx a))))#

#=a xx a xx a xx a xx a = a^5#