How do you simplify #Log 5 = 2 - Log x#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Sihan Tawsik Jan 29, 2016 here is how, Explanation: #log5=2-logx# #or,log5+logx=2# #or,log(5x)=2# #or,10^(log(5x))=10^2# #or,5x=100# #or,x=100/5# #or,x=20# 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 1262 views around the world You can reuse this answer Creative Commons License