How do you solve #log x+log 20=2#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Alan P. Nov 29, 2015 #x=5# Explanation: #log(x)+log(20) = log(20x)# So, if #log(x)+log(20)=2# then #color(white)("XXX")log(20x)=2# #rArrcolor(white)("XX")10^(log(20x)) = 10^2# #rArrcolor(white)("XX")20x = 100# #rArrcolor(white)("XX")x=5# 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 5155 views around the world You can reuse this answer Creative Commons License