How do you solve #6*4^(9x+1)=21#? Precalculus Properties of Logarithmic Functions Natural Logs 1 Answer Alan N. Nov 22, 2016 #x~=-0.01070# Explanation: #6*4^(9x+1)=21# #4^(9x+1)=21/6=7/2# #4^(9x) xx 4 =7/2# #4^(9x) = 7/8# #9xlog4 = log(7/8) = log7 - log8# #9x = (log7-log8)/log4# #9x ~= -0.09632# #x~= -0.01070# 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 1429 views around the world You can reuse this answer Creative Commons License