How do you simplify #2^ 0#?

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Answer 1 · 2016-02-29T19:05:48

#2^0 = 1#

For any non-zero number #a#:

#a^0 = 1#

Note that #0^0# is undefined, except in restricted circumstances.

#color(white)()#
Here's one way of thinking of it:

If #n# is a positive integer, then:

#2*n = overbrace(2+2+...+2)^(n color(white)(x)"times")#

When #n = 0# then #2*n# is an empty sum, with value #0#, the identity under addition.

#2^n = overbrace(2xx2xx...xx2)^(n color(white)(x)"times")#

When #n=0# then #2^n# is an empty product, with value #1#, the identity under multiplication.

Answer 2 · 2016-02-29T19:07:36

anything powered zero is equal to one

Answer 3 · 2016-02-29T19:15:27

1

Think in:
#2^1/2^1=1#

Because we know that a number divided by itself is 1.

And as you may already know, when you divide exponents you just subtract them.
To simplify:
#2^(1-1)=1#
#2^0=1#

There is also a pattern:
#2^3=8 #
#2^2=4#
#2^1=2#
Same base, decrease the exponent by 1 and get half the value
so:
#2^0=1#