How do you simplify #4^n/(3^(n-1))#? Algebra Exponents and Exponential Functions Exponential Properties Involving Quotients 1 Answer smendyka Jul 20, 2017 See a solution process below: Explanation: Multiply the expression by #1# or #3/3#: #3/3 * 4^n/3^(n - 1) = (3 * 4^n)/(3 * 3^(n-1)) = (3 * 4^n)/(3^1 * 3^(n-1)) =# #(3 * 4^n)/(3^(1+n-1)) = (3 * 4^n)/3^n = 3*4^n/3^n = 3(4/3)^n# 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 7815 views around the world You can reuse this answer Creative Commons License