How do you solve #logx = log5 + 3log3#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Bdub Apr 21, 2016 #x=135# Explanation: #log x= log 5 + log 3^3# #log x = log5 + log 27# #log x=log (5*27)# #log x=log 135# #x=135# 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 2335 views around the world You can reuse this answer Creative Commons License