How do you solve #Ln (x-1) + ln (x+2) = 1#? Precalculus Properties of Logarithmic Functions Natural Logs 1 Answer Cesareo R. Jun 20, 2016 #x = 1/2 (-1 +sqrt[9 + 4 e]) = 1.72896# Explanation: #Ln (x-1) + ln (x+2) = 1->Ln(x-1)(x+2)=Ln e# #(x-1)(x+2)-e=0# Solving for #x# #x = 1/2 (-1 pm sqrt[9 + 4 e]) = {-2.72896, 1.72896}# Now, substituting those values in #(x-1)# and #(x+2)# keeping in mind that both must be positive, we get at the solution. #x = 1/2 (-1 +sqrt[9 + 4 e]) = 1.72896# Answer link Related questions What is the natural log of e? What is the natural log of 2? How do I do natural logs on a TI-83? How do I find the natural log of a fraction? What is the natural log of 1? What is the natural log of infinity? Can I find the natural log of a negative number? How do I find a natural log without a calculator? How do I find the natural log of a given number by using a calculator? How do I do natural logs on a TI-84? See all questions in Natural Logs Impact of this question 2620 views around the world You can reuse this answer Creative Commons License