How do you solve for Kc in #7.2647 = -ln(Kc) #? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Trevor Ryan. Mar 30, 2016 #K_c=e^(-7.2647)=0.0007# Explanation: #7.2647=-ln(K_c)# #therefore ln(K_c)=-7.2647# #thereforee^(ln(K_c))=e^(-7.2647)# #therefore K_c=e^(-7.2647)=0.0007#. Answer link Related questions What is a logarithmic model? How do I use a logarithmic model to solve applications? What is the advantage of a logarithmic model? How does the Richter scale measure magnitude? What is the range of the Richter scale? How do you solve #9^(x-4)=81#? How do you solve #logx+log(x+15)=2#? How do you solve the equation #2 log4(x + 7)-log4(16) = 2#? How do you solve #2 log x^4 = 16#? How do you solve #2+log_3(2x+5)-log_3x=4#? See all questions in Logarithmic Models Impact of this question 2954 views around the world You can reuse this answer Creative Commons License