How do you solve #7+3lnx=5#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Jan 19, 2017 #x=e^(-2/3)=0.51342# Explanation: #7+3lnx=5# can be written as #7+3lnx-7=5-7# or #3lnx=-2# or #lnx=-2/3# i.e. #x=e^(-2/3)=0.51342# (using a scientific calculator) 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 5068 views around the world You can reuse this answer Creative Commons License