How do you solve #Log 378 = x#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer ali ergin Mar 7, 2016 #log378=log 2+log 7+3 log 3# Explanation: #log 378=x# #378=2*7*3^3# #log(2*7*3^3)=log 2+log 7+3 log 3# 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 1955 views around the world You can reuse this answer Creative Commons License