How do you solve #-3e^(7a+9)+6=-6#? Precalculus Properties of Logarithmic Functions Natural Logs 1 Answer Shwetank Mauria Oct 8, 2016 #3e^(7a+9)+6==-6# does not have any solution. Explanation: #3e^(7a+9)+6==-6# #hArr3e^(7a+9)==-6-6=-12# and #e^(7a+9)==-12/3=-4# But no power of #e# can be negative, as for all #x's#, #e^x# is positive. graph{e^x [-10, 10, -5, 5]} Hence, #3e^(7a+9)+6==-6# does not have any solution. 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 1637 views around the world You can reuse this answer Creative Commons License