How do you solve #log x + log (x-3)=1#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Vinícius Ferraz Dec 6, 2015 #b = 10 Rightarrow x = (3 + 7)/2 = 5# Explanation: #log_b [x(x-3)] = log_b b# #x(x-3) = b# #x^2 -3x - b = 0# #x = (3 ± sqrt{9 + 4b})/2# Consider #x > 0# and #x - 3 > 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 1057 views around the world You can reuse this answer Creative Commons License