How do you evaluate #(-2)^5#? Prealgebra Exponents, Radicals and Scientific Notation Exponents 1 Answer MeneerNask Jan 11, 2017 In fact, you're looking at #-2# multiplied 5 times by itself Explanation: The #-# signs cancel each other two by two. in this case there are an odd number of #-#'s, so the answer must be negative. Re-write: #(-2)^5=-2^5=-32# 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 7736 views around the world You can reuse this answer Creative Commons License