How do you solve #ln x + ln (x-2) = 1#? Precalculus Properties of Logarithmic Functions Natural Logs 1 Answer Cesareo R. May 20, 2016 #x = 1 + sqrt[1 + e]# Explanation: From #log_e x + log_e(x-2) = log_e(e)# we conclude #log_e(x(x-2))=log_e(e)# and also #x(x-2)=e#. Solving for #x# we obtain #x = 1 + sqrt[1 + e]# 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 1271 views around the world You can reuse this answer Creative Commons License