How do you evaluate #ln (1/0.32)#? Precalculus Properties of Logarithmic Functions Natural Logs 1 Answer Tony B Nov 25, 2016 #1.1394 # to 4 decimal places Explanation: #ln(1/0.32)# is the same as: #ln(1)-ln(0.32) =1.1394 # to 4 decimal places 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 1893 views around the world You can reuse this answer Creative Commons License