How do you solve #log_2 (x+1) = 0#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Aug 8, 2018 #x=0# Explanation: As #log_2(x+1)=0#, we have #2^0=x+1# i.e. #x+1=1# i.e. #x=0# 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 3168 views around the world You can reuse this answer Creative Commons License