How do you solve #log_4(1-x)=1#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer GiĆ³ Jul 22, 2015 I found: #x=-3# Explanation: Use the definition of log as: #log_ax=b ->x=a^b# to get: #1-x=4^1# #x=1-4# #x=-3# 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 1663 views around the world You can reuse this answer Creative Commons License