How do you evaluate #(\frac { 2^ { - 1} \cdot 2^ { - 2} } { 2} ) ^ { 0}#?

Question

531 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-19T14:10:22

1

According to Law of indices, anything raised to the power of 0 #(x^0 = 1)# where x represent any real numbers or integer..

Solution

#((2^-1xx2^-2)/2)^0#

#(2^[-1+(-2)]/2)^0#

#(2^(-1-2)/2)^0#

#(2^(-3)/2)^0#

#(2^(-3)-:2)^0#

Remember that #-># #2=2^1#

#:.# #(2^(-3)-:2^(1))^0#

#(2^(-3-1))^0#

#(2^(-4))^0#

#(1/2^4)^0#

#(1/16)^0#

#(1^0/2^0)#

#(1/1)#

Answer #-># 1