How do you solve #ln 2 + ln x = 5#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer sente Nov 14, 2015 #x = e^5/2# Explanation: We will use the following properties of logarithms: #ln(x) + ln(y) = ln(xy)# #e^(ln(x)) = x# #ln(2) + ln(x) = 5# #=> ln(2x) = 5# (by the first property) #=> e^(ln(2x)) = e^5# #=>2x = e^5# (by the second property) #=> x = e^5/2# 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 9630 views around the world You can reuse this answer Creative Commons License