How do you simplify #2^-4#? Prealgebra Exponents, Radicals and Scientific Notation Exponents 1 Answer Shwetank Mauria Jun 29, 2016 #2^(-4)=1/16# Explanation: As #a^m-:a^n=a^(m-n)#, if #m=0#, we have #a^0-:a^n=a^(0-n)# or #1/a^n=a^(-n)# i.e. #a^(-n)=1/a^n# Hence, #2^(-4)=1/2^4=1/(2xx2xx2xx2)=1/16# Answer link Related questions How do you simplify #c^3v^9c^-1c^0#? How do you simplify #(- 1/5)^-2 + (-2)^-2#? How do you simplify #(4^6)^2 #? How do you simplify #3x^(2/3) y^(3/4) (2x^(5/3) y^(1/2))^3 #? How do you simplify #4^3ยท4^5#? How do you simplify #(5^-2)^-3#? How do you simplify and write #(-5.3)^0# with positive exponents? How do you factor #12j^2k - 36j^6k^6 + 12j^2#? How do you simplify the expression #2^5/(2^3 times 2^8)#? When can I add exponents? See all questions in Exponents Impact of this question 2003 views around the world You can reuse this answer Creative Commons License