How do you solve #log_b(3) = .234#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Alan P. Apr 18, 2016 #b~~109.39# Explanation: If #log_b (3) = .234# then (by definition of #log# #color(white)("XXX")b^.234 = 3# #color(white)("XXX")(b^.234)^(1/.234)=3^(1/.234)# #color(white)("XXX")b=3^(1/.234)~~3^4.2735~~109.3905# 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 1763 views around the world You can reuse this answer Creative Commons License