How do you simplify #((2r^3t^6)/(5u^9))^4#? Prealgebra Exponents, Radicals and Scientific Notation Exponents 1 Answer Shwetank Mauria Aug 23, 2017 #((2r^3t^6)/(5u^9))^4=(16r^12t^24)/(625u^36)# Explanation: #((2r^3t^6)/(5u^9))^4# = #(2^4r^(3xx4)t^(6xx4))/(5^4u^(9xx4))# = #(16r^12t^24)/(625u^36)# 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 2308 views around the world You can reuse this answer Creative Commons License