How do you solve #25^(2x+1) =144#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer A. S. Adikesavan Apr 27, 2016 #x=(log 144/log 25 - 1)/2=(log 12/log 5 - 1)/2#. Explanation: Equate logarithms. #(2x+1)log 25=log 144#. #(2x+1)=log 144/log 25#. #x=(log 144/log 25 - 1)/2# 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 3227 views around the world You can reuse this answer Creative Commons License