Question #04063 Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Cesareo R. Mar 3, 2017 #x=1/0.3 log(111)# Explanation: #100/1000=1/10=1/(1+999e^(-0.3x))# so #1+999e^(-0.3x)=10->e^(-0.3x)=9/999=1/111# finally #-0.3x=log(1/111)=-log(111)->x=1/0.3 log(111)# 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 959 views around the world You can reuse this answer Creative Commons License