How do you solve #2^x=4^(x+1)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer George C. Aug 22, 2015 Re-express the right hand side as #2^(2x+2)# to find #x = -2# Explanation: #(a^b)^c = a^(bc)# and #4 = 2^2#, so... #2^x = 4^(x+1) = (2^2)^(x+1) = 2^(2(x+1)) = 2^(2x+2)# So #x = 2x+2# Subtract #(x + 2)# from both sides to get #x = -2# 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 1060 views around the world You can reuse this answer Creative Commons License