How do you evaluate #3^ln0.2#? Precalculus Properties of Logarithmic Functions Natural Logs 1 Answer Shwetank Mauria Feb 24, 2017 #3^(ln0.2)=0.171# Explanation: Let #3^(ln0.2)=x#. Taking natural log on both sides, we get #ln0.2ln3=lnx# and hence #x=e^(ln0.2ln3)# = #e^(-1.609xx1.0986)# = #e^(-1.76815)=0.171# 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 2510 views around the world You can reuse this answer Creative Commons License