How do you solve #3^(2x+1) = 5^(x+2)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer A. S. Adikesavan May 19, 2016 #x=(2 log 5 - log 3)/(2 log 3 - log 5 )=3.607#, nearly. Explanation: Equating logarithms, #(2x+1)log 3=(x+2)log 5# So, #x=(2 log 5 - log 3)/(2 log 3 - log 5 )=3.607#, nearly. 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 7598 views around the world You can reuse this answer Creative Commons License