How do you solve #7^(x – 2) = 12^x#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer A. S. Adikesavan Apr 12, 2016 #x=(2 log 7)/(log 7-log 12)=-7.22#, nearly. Explanation: Use #log (a^m)=m log a#. Equating the logarithms, #(x-2) log 7=x log 12#. So, #x (log 7-log 12)=2 log 7. x=(2 log 7)/(log 7-log 12)# 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 1072 views around the world You can reuse this answer Creative Commons License