How do you solve #lnx+ln3x=12#? Precalculus Properties of Logarithmic Functions Natural Logs 1 Answer A. S. Adikesavan Nov 19, 2016 #+-e^6/sqrt3=+-232.92#, nearly Explanation: #ln x + ln (3x) = ln (x()(3x))=ln(3x^2)#. Inversely, #3x^2=e^12#. So, #x=+-sqrt(e^12/3)# #=+-e^6/sqrt3#. 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 3326 views around the world You can reuse this answer Creative Commons License