How do you solve #log_2x + log_2(x-2) = 3#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Bdub Apr 21, 2016 #x=4# Explanation: Use Property: #log_b(xy)=log_bx+log_by and log_by=x iff x^b=y# #log_2(x(x-2))=3# #2^3=x^2-2x# #8=x^2-2x# #0=x^2-2x-8# #0=(x-4)(x+2)# #x-4=0 or x+2=0# #x=4 or x=-2# #x=4# 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 19716 views around the world You can reuse this answer Creative Commons License