How do you solve #5^(2x-1) = 7^(2x)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer VinÃcius Ferraz Jul 3, 2018 #x = frac{log 5}{2(log 5 - log 7)}# Explanation: #log 5^{2x - 1} = log 7^{2x}# #(2x - 1) log 5 = 2x log 7# #x (2 log 5 - 2 log 7) = 1 log 5# #x = frac{log 5}{log 25 - log 49}# #(25/49)^x = 5# 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 3441 views around the world You can reuse this answer Creative Commons License