How do you simplify the expression #2^5/(2^3 times 2^8)#? Prealgebra Exponents, Radicals and Scientific Notation Exponents 1 Answer Shwetank Mauria Jun 21, 2016 #2^5/(2^3xx2^8)=1/2^6# Explanation: #2^5/(2^3xx2^8)# = #2^5/(2^3xx2^(3+5))# = #2^5/(2^3xx2^3xx2^5)# = #(1cancel2^5)/(2^3xx2^3xxcancel2^5)# = #1/(2^3xx2^3)=1/2^(3+3)=1/2^6# 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#? When can I add exponents? What is the Zero Exponent Rule? See all questions in Exponents Impact of this question 3299 views around the world You can reuse this answer Creative Commons License