How do you solve #-8.7e^(6.6-6x)-7=-32.8#? Precalculus Properties of Logarithmic Functions Natural Logs 1 Answer Ratnaker Mehta Aug 4, 2016 #x=0.9188# Explanation: Let us rewrite the given eqn., by removing all #-ve# signs, as # 8.7e^(6.6-6x)+7=32.8# # rArr 8.7e^(6.6-6x)=25.8# # rArr e^(6.6-6x)=25.8/8.7# # rArr log_10{e^(6.6-6x)}=log_10 25.8-log_10 8.7# # rArr (6.6-6x)log_10e=1.4116-0.9395=0.4721# # rArr (6.6-6x)(0.4343)=0.4721# # rArr (6.6-6x)=0.4721/0.4343=1.087# # rArr (6.6-1.087)/6=x=0.9188#. 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 1285 views around the world You can reuse this answer Creative Commons License