How do you solve #2^(x+1) = 3^x#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer A08 Jan 23, 2016 #xapprox1.7093# Explanation: #2^(x+1)=3^x# Taking log of both sides #log2^(x+1)=log3^x# #implies (x+1)log2=xlog3# #implies (x+1)xx0.3010=x xx0.4771# Rearranging we get # (-0.3010+0.4771)x=0.3010# #implies x=0.3010/(0.4771-0.3010)# #xapprox1.7093# 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 1176 views around the world You can reuse this answer Creative Commons License