How do you solve #4x^(2/3) - 5 = 20#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer José F. Apr 17, 2016 #x=125/8# Explanation: #4x^(2/3)-5=20# #4x^(2/3)=25# #x^(2/3)=25/4# #x=(25/4)^(3/2)# #x=(5^2/2^2)^(3/2)# #x=(5/2)^(3)# #x=125/8# Answer link Related questions What is a logarithmic model? How do I use a logarithmic model to solve applications? What is the advantage of a logarithmic model? How does the Richter scale measure magnitude? What is the range of the Richter scale? How do you solve #9^(x-4)=81#? How do you solve #logx+log(x+15)=2#? How do you solve the equation #2 log4(x + 7)-log4(16) = 2#? How do you solve #2 log x^4 = 16#? How do you solve #2+log_3(2x+5)-log_3x=4#? See all questions in Logarithmic Models Impact of this question 1237 views around the world You can reuse this answer Creative Commons License