How do you simplify #a^-2times a^3timesa^4#?
1 Answer
Explanation:
If
#a^n = overbrace(a xx a xx .. xx a)^"n times"#
So if
#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^(-n) = 1/underbrace(a xx a xx .. xx a)_"n times"#
In fact, if
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#