How do you solve # Log 5 = 2 - Log x#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer José F. Feb 27, 2016 #x=e^2/5# Explanation: #log5=2−logx# #logx=2−log5# #logx= log (e^2)-log5# #logx= log (e^2/5)# #x=e^2/5# 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 2129 views around the world You can reuse this answer Creative Commons License