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 1099 views around the world You can reuse this answer Creative Commons License