How do you solve #log_3 (2x-1) = 3#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer mason m Dec 25, 2015 #x=14# Explanation: To undo a logarithm with base #3#, exponentiate both sides with a base of #3#. #3^(log_3(2x-1))=3^3# #2x-1=27# #2x=28# #x=14# 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 2596 views around the world You can reuse this answer Creative Commons License