How do you simplify #(4^6)^2 #? Prealgebra Exponents, Radicals and Scientific Notation Exponents 1 Answer Shwetank Mauria Jun 8, 2016 #(4^6)^2=4^12=16777216# Explanation: WE use two simple formulas here - one #a^n=axxaxxaxx....xxa#, where #a# is multiplied #n# times ant two #(a^m)^n=a^(mn)# Hence, #(4^6)^2=4^(6xx2)=4^12# = #4xx4xx4xx4xx4xx4xx4xx4xx4xx4xx4xx4=16777216# 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 #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? What is the Zero Exponent Rule? See all questions in Exponents Impact of this question 4048 views around the world You can reuse this answer Creative Commons License