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 1088 views around the world You can reuse this answer Creative Commons License