How do you solve #log(x-3)+log x=1#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Bdub · EZ as pi May 4, 2016 #x=5 # Explanation: Use Properties: #log_b(xy)=log_b x+log_by# #log_bx=y iff b^y=x# #log (x(x-3))=1# #color(white)(xxxxxx)# [ 1 = log10] #log(x^2-3x)=log10# #x^2-3x^1 = 10^1# #x^2-3x-10=0# #(x-5)(x+2)=0# #x=5 or x=-2# 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 4834 views around the world You can reuse this answer Creative Commons License