How do you solve #z^(-1/4) = 1/3#? Algebra Exponents and Exponential Functions Negative Exponents 1 Answer 256 Jan 18, 2017 #z^(-1/4)=1/3 <=> z=81# Explanation: #z^(-1/4)=1/3# #<=># First since #x^(-1)=1/x#, and #x^(ab)=(x^a)^b# #z^((-1)(1/4))=(z^(1/4))^(-1)=1/z^(1/4)=1/3# #<=># #1=z^(1/4)/3# multiply both sides by #z^(1/4)# #<=># #3=z^(1/4)# multiply both sides by 3 #<=># #3^4=(z^(1/4))^4# raise both sides to the power of 4 #<=># #3^4=z# #<=># #(3^2)^2=(3^2)(3^2)=9(9)=ul(81=z)# Answer link Related questions What are Negative Exponents? How are negative exponents used in real life? How do negative exponents represent repeated division? How does a negative exponent affect the base number? How do you simplify expressions with negative exponents? How do you evaluate expressions with negative exponents? How do negative exponents affect fractions? Why are negative exponents used? What is the exponent of zero property? How do you rewrite the expression #\frac{x^2}{y^3}# without fractions? See all questions in Negative Exponents Impact of this question 381 views around the world You can reuse this answer Creative Commons License