How do you solve #10^(-12x)+6=100#? Precalculus Properties of Logarithmic Functions Natural Logs 1 Answer EZ as pi Jul 16, 2016 #x = -0.1644# Explanation: #10^(-12x)+6=100# #10^(-12x)=94# As the variable is in the index, we will use logs. #log10^(-12x)=log94# #-12x xxlog10=log94 " "log 10 = 1# #x =log94/(-12)# #x = -0.1644# 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 2068 views around the world You can reuse this answer Creative Commons License