How do you simplify #(n^-1)^4/n^4# and write it using only positive exponents? Algebra Exponents and Exponential Functions Exponential Properties Involving Quotients 1 Answer Tony B Aug 22, 2016 #1/n^8# Explanation: In the same way that #2/3 = 2xx1/3" "# split #" "((n^(-1))^4)/n^4# #(n^(-1))^4xx1/n^4# But #" "n^(-1) -> 1/n# so #(n^(-1))^4 =( 1/n)^4 = 1/n^4# Hence: #((n^(-1))^4)/n^4" " =" " 1/n^4xx1/n^4" "=" "1/n^8# Answer link Related questions What is the quotient of powers property? How do you simplify expressions using the quotient rule? What is the power of a quotient property? How do you evaluate the expression #(2^2/3^3)^3#? How do you simplify the expression #\frac{a^5b^4}{a^3b^2}#? How do you simplify #((a^3b^4)/(a^2b))^3# using the exponential properties? How do you simplify #\frac{(3ab)^2(4a^3b^4)^3}{(6a^2b)^4}#? Which exponential property do you use first to simplify #\frac{(2a^2bc^2)(6abc^3)}{4ab^2c}#? How do you simplify #(x^5y^8)/(x^4y^2)#? How do you simplify #[(2^3 *-3^2) / (2^4 * 3^-2)]^2#? See all questions in Exponential Properties Involving Quotients Impact of this question 1429 views around the world You can reuse this answer Creative Commons License