How do you solve 4^(x+4) = 8^x4x+4=8x? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer GiĆ³ Jul 23, 2015 I found: x=8x=8 Explanation: You can write it as: (2^2)^(x+4)=(2^3)^x(22)x+4=(23)x 2^(2(x+4))=2^(3x)22(x+4)=23x take the log_2log2 on both sides to get rid of the 22 (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 1779 views around the world You can reuse this answer Creative Commons License