How do you solve #4^(x+2)=48#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Bdub Oct 27, 2016 #x~~0.7925# Explanation: #4^(x+2) = 48# #ln 4^(x+2) = ln 48# #(x+2) ln 4 = ln 48# #x+2=ln 48/ln 4# #x=ln 48/ln 4-2# #x~~0.7925# 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 2599 views around the world You can reuse this answer Creative Commons License