How do you solve #4^x-3=5#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Gerardina C. Jul 20, 2016 #x=3/2# Explanation: First let's have the equivalent equation #4^x=8# Then, since #4=2^2 and 8=2^3#, let's substitute: #2^(2x)=2^3# so #2x=3# #x=3/2# 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 900 views around the world You can reuse this answer Creative Commons License