How do you simplify #x^-1/(4x^4)# and write it using only positive exponents? Algebra Exponents and Exponential Functions Exponential Properties Involving Quotients 1 Answer LM Apr 4, 2018 #1/(4x^5)# Explanation: #(x^-1)/(4x^4)# can be written as #x^(-1) * 1/(4x^4)#. #x^-1# is the same as #1/(x^1)#, which is #1/x#. this means that #(x^-1)/(4x^4) = 1/x * 1/(4x^4)# #1/x * 1/(4x^4) = 1/(x * 4x^4)# #x * 4x^4 = 4x^5# #1/(x * 4x^4) = 1/(4x^5)#. hence, #(x^-1)/(4x^4) = 1/(4x^5)#. 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 6455 views around the world You can reuse this answer Creative Commons License