How do you solve #-2.3*e^(-6m-10)=-38.8#? Precalculus Properties of Logarithmic Functions Natural Logs 1 Answer A. S. Adikesavan Jan 25, 2017 #m =- 1/6(10+ln(38.8/2.3))=-2.1376#, nearly. Explanation: Here, #e^(-(6m+10))=38.8/2.3#. Inversely, -#6m-10=ln(38.8/2.3)#, giving #m= -1/6(10+ln(38.8/2.3))# 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 495 views around the world You can reuse this answer Creative Commons License