How do you solve #logx+log5=1#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Alan N. Apr 15, 2016 #x = 2# (Assuming #Log# is #Log_ 10#) Explanation: #Log(x) + Log(5) = 1# Applying #Log(a) + Log(b) = Log(ab)# # -> Log(5x) = 1# Assuming #Log# is #Log_10# #-> 5x = 10^1# #5x = 10# #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 5874 views around the world You can reuse this answer Creative Commons License