How do you simplify #(m^7n^4p)/(m^3n^3p)#? Algebra Exponents and Exponential Functions Exponential Properties Involving Quotients 1 Answer Azimet Jan 18, 2017 #m^4n# (if #m, n,p,# are all non-zero) Explanation: We know that #a^m/a^n = a^(m-n)#, for #a!=0#. So, provided #m, n,p,# are all non-zero, the result is: #m^7/m^3 n^4/n^3 p/p = m^4n# 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 2090 views around the world You can reuse this answer Creative Commons License