How do you solve #Log(x)-2=log(x-4)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer José F. Mar 28, 2018 x=4/99 Explanation: #log(x)-2=log(x-4)# #x/10^2=x-4,iff x>0 and x>4# #x/10^2-x=-4# #x-100x=-400# #-99x=-400# #x=400/99# x>4 so the solution is correct. 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 2449 views around the world You can reuse this answer Creative Commons License