How do you solve #4^(x+4) = 8^x#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer GiĆ³ Jul 23, 2015 I found: #x=8# Explanation: You can write it as: #(2^2)^(x+4)=(2^3)^x# #2^(2(x+4))=2^(3x)# take the #log_2# on both sides to get rid of the #2# (where you have #cancel(log_2)(cancel(2)^x)=x#) and get: #2(x+4)=3x# #2x+8=3x# #x=8# 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 1509 views around the world You can reuse this answer Creative Commons License