How do you simplify #2^1000 - 2^999#?

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Answer 1 · 2018-05-29T14:57:43

#2^999#

We need to remember that:
#n^a xx n^b = n^(a+b)#

#n^a / n^b = n^(a-b)#

So here's what we gonna do

#2^1000 - 2^999#

#= (2) (2^999) - (1) (2^999)#

#= (2-1) (2^999)#

#= (2^999)#

Answer 2 · 2018-05-29T15:03:26

#x=2^999#

We take,

#x=2^(1000)-2^999#

#x=2^(999+1)-2^999#

#x=2^999*2^1-2^999...to[becausea^(m+n)=a^m*a^n]#

#x=2^999(2^1-1)#

#x=2^999(1) #

#x=2^999#