How do you solve #ln(x+4)=ln7#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Alexander Jul 16, 2016 Taking the #e#-xponential of both sides we get #cancel(e)^(cancel(ln)(x+4)) = cancel(e)^(cancel(ln) 7)# #x+4 = 7# Subtracting #4# from both sides gives us our final answer of #x = 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 5026 views around the world You can reuse this answer Creative Commons License