Question #85638 Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer A. S. Adikesavan Apr 29, 2016 #x =-(log 3 - log 2 )/log 3=-0.3691#, nearly. Explanation: Use# (ab)^n=a^nb^n, a^(m+n)=a^ma^n and log(a^n)=n log a# #((3)(2))^(x+1)=3^(x+1)2^(x+1)=2^(x+2)#. So, #3^(x+1)=2#. Equating logarithms and simplifying, #x =-(log 3 - log 2 )/log 3=-0.3691#, nearly. 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 958 views around the world You can reuse this answer Creative Commons License